23.22 - Solved Examples on Definite Integral as Limit of Sum

In the previous section, we saw the definite integral as limit of sum. We saw a solved example also. In this section, we will see a few more solved examples.

Solved Example 23.70
Evaluate $\small{\int_0^5{\left[x+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^5{\left[x+1 \right]dx}}$ is the area bounded by the four items:
   ♦ The line y = f(x) = x+1.
   ♦ The vertical line x = 0
   ♦ The vertical line x = 5
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.5 below. Our task is to find the area of the violet shaded area.

Fig.23.5

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 5
So $\small{h~=~\frac{b-a}{n}~=~\frac{5-0}{n}~=~\frac{5}{n}}$

3. Now the formula becomes:
$\small{\int_0^5{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{5-0}{n} \left[f(a+\frac{5}{n})~+~f(a+\frac{10}{n})~+~~.~.~.~+f(a+\frac{5n}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{5}{n} \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$

4. Let us determine the quantity inside the square brackets:

$\small{ \left[(a+\frac{5}{n}\,+1)~+~(a+\frac{10}{n}\,+1)~+~.~.~.~+(a+\frac{5n}{n}+1) \right]}$

5. So we have to do three summations:

(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}~=~na~=~n(0)~=~0}$

(ii) $\small{\frac{5}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{5}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{5(n+1)}{2}~=~\frac{5n}{2}~+~\frac{5}{2}}$

(Here we use the technique of arithmetic progression)

(iii) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n(1)~=~n}$

6. So the total of three summations is:

$\small{0~+~\frac{5n}{2}~+~\frac{5}{2}~+~n}$

7. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{5}{n} \left[\frac{5n}{2}~+~\frac{5}{2}~+~n \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \left[\frac{25}{2}~+~\frac{25}{2n}~+~5 \right]~=~\frac{35}{2}}$

Solved Example 23.71
Evaluate the definite integral $\small{\int_0^2{\left[x^2+1 \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[x^2+1 \right]dx}}$ is the area bounded by the four items:
   ♦ The line y = f(x) = x²+1.
   ♦ The vertical line x = 0
   ♦ The vertical line x = 2
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.6 below. Our task is to find the area of the violet shaded area.

Fig.23.6

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \bigg[\big[(a+\frac{2}{n})^2~+~1\big]~+~\big[(a+\frac{4}{n})^2~+~1 \big]~+~\big[(a+\frac{6}{n})^2~+~1 \big]~+~~.~.~.}$

$\small{~~~~~~~~~~.~.~.~+~\big[(a+\frac{2n}{n})^2~+~1 \big] \bigg]}$

4. Let us determine the quantity inside the large square brackets:

$\small{~=~ \bigg[\big[a^2 + \frac{4a}{n}+\frac{2^2}{n^2}~+~1\big]~+~\big[a^2 + \frac{8a}{n}+\frac{4^2}{n^2}~+~1 \big]~+~\big[a^2 + \frac{12a}{n}+\frac{6^2}{n^2}~+~1 \big]~+~ }$

$\small{~~~~~~~~.~.~.~+~\big[a^2 + \frac{2(2an)}{n}+\frac{2^2\,n^2}{n^2}~+~1 \big] \bigg]}$

5. So we have to do four summations:

(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(0^2)~=~0}}$

(ii) $\small{\frac{4a}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)~~~~~=~0, ~~~\because a~=~0}$

(iii) $\small{\frac{2^2}{n^2}\left(1^2~+~2^2~+~3^2~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{2^2}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{4n}{3}~+~2~+~\frac{2}{3n}}$

(Here we use the technique that we saw in section 9.5)

(iv) $\small{1~+~1~+~1~+~.~.~.~ \text{n terms}~=~n (1)~=~n}$

6. So the total of 4 summations is:

$\small{0+0+\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n}$

8. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \left[\frac{4n}{3}~+~2~+~\frac{2}{3n}~+~n \right]}$

$\small{~=~\lim_{n\rightarrow \infty}  \left[\frac{8}{3}+\frac{4}{n}+\frac{4}{3n^2}+2 \right]~=~\frac{8}{3}+2~=~\frac{14}{3}}$

Solved Example 23.72
Evaluate the definite integral $\small{\int_0^2{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_0^2{\left[e^x \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = e^x.
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 2
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.7 below. Our task is to find the area of the violet shaded area.

Fig.23.7

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 2
So $\small{h~=~\frac{b-a}{n}~=~\frac{2-0}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2-0}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$

4. Let us determine the quantity inside the large square brackets:

$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$

$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$

Here $\small{e^a~=~1~~\because a = 0}$

5. The n terms inside the square brackets, form a geometric progression.

• First term = $\small{e^{\frac{2}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$

6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$

7. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n} \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} \left({e^{2}}~-~1 \right)}{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\frac{e^{0} \left({e^{2}}~-~1 \right)}{1} ~=~e^2 ~-~1}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

Solved Example 23.73
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{-1}^{1}{\left[e^x \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = e^x.
   ♦ The vertical line x = −1
   ♦ The vertical line x = 1
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.8 below. Our task is to find the area of the violet shaded area.

Fig.23.8

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = -1 and b = -1

So $\small{h~=~\frac{b-a}{n}~=~\frac{1-(-1)}{n}~=~\frac{2}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \left[f(a+\frac{2}{n})~+~f(a+\frac{4}{n})~+~f(a+\frac{6}{n})~+~~.~.~.~+f(a+(n)\frac{2}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n} \big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.~+~e^{a+\frac{2n}{n}}\big]}$

4. Let us determine the quantity inside the large square brackets:

$\small{\big[e^{a+\frac{2}{n}}~+~e^{a+\frac{4}{n}}~+~e^{a+\frac{6}{n}}~+~~.~.~.+~e^{a+\frac{2n}{n}}\big]}$

$\small{~=~e^a\big[e^{\frac{2}{n}}~+~e^{\frac{4}{n}}~+~e^{\frac{6}{n}}~+~~.~.~.+~e^{\frac{2n}{n}}\big]}$

Here $\small{e^a~=~\frac{1}{e}~~\because a = -1}$

5. The n terms inside the square brackets, form a geometric progression.

• First term = $\small{e^{\frac{2}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{4}{n}}}{e^{\frac{2}{n}}}~=~\frac{e^{\frac{6}{n}}}{e^{\frac{4}{n}}}~=~e^{\frac{2}{n}}}$

6. Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^n~-~1 \right)}{r~-~1}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{\frac{2n}{n}}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}~=~\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1}}$

7. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{1}{e} \right) \big[\frac{e^{\frac{2}{n}}~\times~\left({e^{2}}~-~1 \right)}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{2}{n}\left(\frac{{e^{2}}~-~1}{e} \right) \big[\frac{e^{\frac{2}{n}}}{e^{\frac{2}{n}}~-~1} \big]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)~\lim_{n\rightarrow \infty}  \Bigg[\frac{e^{\frac{2}{n}} }{\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}} \Bigg]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)  \Big[\frac{1}{1} \Big]}$

$\small{~=~\left(\frac{{e^{2}}~-~1}{e} \right)  ~=~e~-~\frac{1}{e}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{2}{n}} \right]~=~e^{\frac{2}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{2}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{2}{n}}~-~1}{\frac{2}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

Solved Example 23.74
Evaluate the definite integral $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ as the limit of a sum.
Solution:
1. $\small{\int_{0}^{4}{\left[x+e^{2x} \right]dx}}$ is the area bounded by the four items:
   ♦ The curve y = f(x) = x+e^2x.
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 4
   ♦ The horizontal line y = 0 (x-axis)

• A rough sketch is shown in the fig.23.9 below. Our task is to find the area of the violet shaded area.

Fig.23.9

2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 4

So $\small{h~=~\frac{b-a}{n}~=~\frac{4-0}{n}~=~\frac{4}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \left[f(a+\frac{4}{n})~+~f(a+\frac{8}{n})~+~f(a+\frac{12}{n})~+~~.~.~.~+f(a+(n)\frac{4}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \bigg[\big[a+\frac{4}{n}~+~e^{2(a+\frac{4}{n})}\big]~+~\big[a+\frac{8}{n}~+~e^{2(a+\frac{8}{n})}\big]~+~\big[a+\frac{12}{n}~+~e^{2(a+\frac{12}{n})}\big]~+~~.~.~.~}$

$\small{~~~~~~~~~.~.~.~+~\big[a+(n)\frac{4}{n}~+~e^{2(a+(n)\frac{4}{n})}\big]\bigg]}$

4. Let us determine the quantity inside the large square brackets. We have to do three summations:

(i) $\small{a~+~a~+~a~+~.~.~.~ \text{n terms}}$

$~=~\small{n a~=~n(0)~=~0}$

(ii) $\small{\frac{4}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{4}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{4(n+1)}{2}~=~2n~+~2}$

(Here we use the technique of arithmetic progression)

(iii) $\small{e^{2(a+\frac{4}{n})}~+~e^{2(a+\frac{8}{n})}~+~e^{2(a+\frac{12}{n})}~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2a}~\times~e^{\frac{8}{n}}~+~e^{2a}~\times~e^{\frac{16}{n}}~+~e^{2a}~\times~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2a}\left(e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~=~e^{\frac{8}{n}}~+~e^{\frac{16}{n}}~+~e^{\frac{24}{n}}~+~.~.~.~ \text{n terms} }$

$\small{~~~~~~\because e^{2a}~=~e^{2(0)}~=~e^0~=~1}$

• So we have a geometric progression.

• First term = $\small{e^{\frac{8}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{16}{n}}}{e^{\frac{8}{n}}}~=~\frac{e^{\frac{24}{n}}}{e^{\frac{16}{n}}}~=~e^{\frac{8}{n}}}$

• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$

• So we get:

Sum of all terms of the G.P

$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{8}{n}}\right)^{n-1}~=~e^{\frac{8n-8}{n}}~=~e^{8-\frac{8}{n}}}$

• So we get:

Sum of all terms inside the brackets

$\small{~=~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1}}$

5. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{4}{n} \big[0~+~2n\,+\,2~+~\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{4}{n} \big[2n\,+\,2 \big]~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~8~+~\lim_{n\rightarrow \infty} \frac{4}{n} \big[\frac{e^{\frac{8}{n}}~\times~\left({e^{8-\frac{8}{n}}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]}$

6. In the above expression, we need to evaluate the limit for the second term. It can be done as follows:

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}}~\times~\left(\frac{e^8}{e^\frac{8}{n}}~-~1 \right)}{e^{\frac{8}{n}}~-~1} \big]~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8~-~e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^8 }{e^{\frac{8}{n}}~-~1} \big]~-~\lim_{n\rightarrow \infty} \frac{8}{2n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{1 }{e^{\frac{8}{n}}~-~1} \big]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty} \frac{8}{n} \big[\frac{e^{\frac{8}{n}} }{e^{\frac{8}{n}}~-~1} \big]}$

$\small{~=~\frac{e^8}{2}\,\lim_{n\rightarrow \infty}  \bigg[\frac{1 }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \bigg]~-~\frac{1}{2}\,\lim_{n\rightarrow \infty}  \big[\frac{e^{\frac{8}{n}} }{\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}} \big]}$

$\small{~=~\frac{e^8}{2}\,  \bigg[\frac{1}{1} \bigg]~-~\frac{1}{2}\, \big[\frac{1}{1} \big]~=~\frac{e^8~-~1}{2}}$

7. So the limit in (3) becomes:

$\small{8~+~\frac{e^8~-~1}{2}~=~\frac{16~+~e^8~-~1}{2}~=~\frac{15~+~e^8}{2}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{8}{n}} \right]~=~e^{\frac{8}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{8}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{8}{n}}~-~1}{\frac{8}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$

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The link below gives a few more solved examples:

Exercise 23.8

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We have completed a discussion on the definite integral as limit of sum. In the next section, we will see fundamental theorem of calculus.

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23.21 - Definite Integrals

In the previous section, we completed a discussion on three standard integrals. In this section, we will see definite integrals.

Let us first see a method to calculate the area of a region. The region that we are considering here, is a special kind of region because, one of it's sides is not a straight line. The method can be written in 15 steps:

1. In fig.23.2 below, the red curve is the graph of y = f(x).

Fig.23.2

• We assume that, for all input x values, f(x) is non negative. So the graph is above the x-axis.
• There are two vertical lines: x = a and x = b. They are drawn in green color.
• So we have a region bounded by the four items below:
   ♦ The curve y = f(x)
   ♦ The vertical line x = a
   ♦ The vertical line x = b
   ♦ The horizontal line y = 0 (x-axis)
• Our aim is to find the area of this region.

2. In fig.23.3 below, the region is divided into strips.

Fig.23.3

• The strips have the following 4 properties:
(i) All strips are formed by vertical lines.
(ii) All strips have the same width. We will denote this width as 'h'.
(iii) There is a total of n strips.
(iv) Each of the n strips, has it's own position.
   ♦ The left edge of the first strip is at x = x₀ = a
   ♦ The left edge of the second strip is at x = x₁
   ♦ The left edge of the third strip is at x = x₂
   ♦ The left edge of the fourth strip is at x = x₃
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the r^th strip is at x = x_r−1
   ♦ The left edge of the (r+1)^th strip is at x = x_r
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ The left edge of the n^th strip is at x = x_n−1
   ♦ The right edge of the n^th strip is at x = x_n = b

3. Based on the above properties of the strips, we can write:
   ♦ x₀ = a
   ♦ x₁ = a+h
   ♦ x₂ = a+h+h = a+2h
   ♦ x₃ = a+2h+h = a+3h
   ♦ x₄ = a+3h+h = a+4h
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_r = a+rh
   ♦ -    -    -    -
   ♦ -    -    -    -
   ♦ x_n = a+nh
4. Now we can write an expression for n.
• We have $\small{x_n~=~a+nh~=~b}$
$\small{\Rightarrow nh~=~b-a}$
$\small{\Rightarrow n~=~\frac{b-a}{h}}$

5. From the above expression, we get an important information:
• h is in the denominator. So when n increases, h decreases.
• Consequently, when n is close to infinity, h will be close to zero.

6. Now select any convenient strip as the r^th strip. In fig.23.3, it is named as ABCD.
• Note that, none of the strips is a rectangle. This is because, the top edge of any selected strip is not horizontal. Our present strip ABCD is also not a rectangle because, the top edge CD is not horizontal.

7. Since CD is not horizontal, we draw two new horizontal lines:
   ♦ Horizontal line CL through C
   ♦ Horizontal line DM through D
• Now we have three regions:
   ♦ Rectangle ABDM
   ♦ Region ABDCA
   ♦ Rectangle ABLC

8. Let us compare the areas of the three regions. From the fig., it is clear that:
• Area of region ABDCA is less than the area of the rectangle ABDM
• Area of region ABDCA is greater than the area of the rectangle ABLC

9. Consider the situation when n is close to infinity.
• In such a situation,
   ♦ h will be very close to zero.
   ♦ B will be very close to A
   ♦ L will be very close to C
   ♦ Segment MD will be very close to segment CL
• In such a situation, all three areas mentioned in (8), can be considered to be equal.

10. It follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABLC.
• Area of rectangle ABLC = AC  × AB = f(x_r-1)  × h
• Here we are using the lower rectangle ABLC to get the area of the region ABCDA. Length of the rectangle is left edge AC.

11. Similarly, it follows that:
• When n is close to infinity, the area of region ABCDA can be calculated easily because, it is equal to the area of rectangle ABDM.
• Area of rectangle ABDM = BD  × AB = f(x_r)  × h
• Here we are using the upper rectangle ABDM to get the area of the region ABCDA. Length of the rectangle is right edge BD.

12. The information in both (10) and (11) are applicable to all strips.
• Let us first apply the information in (10).
   ♦ Area of first strip = left edge × base = f(x₀) × h
   ♦ Area of second strip = left edge × base = f(x₁) × h
   ♦ Area of third strip = left edge × base = f(x₂) × h
   ♦ Area of fourth strip = left edge × base = f(x₃) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = left edge × base = f(x_r−1) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = left edge × base = f(x_n−1) × h
• So total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]= $\small{h\sum\limits_{r=0}^{r=n-1}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
Total area of all n strips
= h[f(x₀) + f(x₁) + f(x₂) + . . . +f(x_n−1)]
= h[f(a) + f(a+h) + f(a+2h) + . . . +f(a+(n−1)h)]
= $\small{h\sum\limits_{k=0}^{k=n-1}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a)~+~f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+(n-1)h) \right]}$

13. Let us now apply the information in (11).
   ♦ Area of first strip = right edge × base = f(x₁) × h
   ♦ Area of second strip = right edge × base = f(x₂) × h
   ♦ Area of third strip = right edge × base = f(x₃) × h
   ♦ Area of fourth strip = right edge × base = f(x₄) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of r^th strip = right edge × base = f(x_r) × h
   ♦ -    -    -    -
   ♦ -    -    -    -   
   ♦ Area of n^th strip = right edge × base = f(x_n) × h
• So total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)] = $\small{h\sum\limits_{r=1}^{r=n}{f(x_r)}}$
• Based on the result in (3), we can write x values in terms of a and h. We get:
• Total area of all n strips
= h[f(x₁) + f(x₂) + f(x₃) + . . . +f(x_n)]
= h[f(a+h) + f(a+2h) + f(a+3h) + . . . +f(a+nh)]
= $\small{h\sum\limits_{k=1}^{k=n}{f(a+kh)}}$
• When n is close to infinity, h will be close to zero. In such a situation, this sum will be equal to the area of PRSQP.
• That is.,
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

14. Let us compare the methods.
• Step (12) gives us a method to find the area of PRSQP.
It uses the left edge of each rectangle.
• Step (13) also gives us a method to find the area of PRSQP.
It uses the right edge of each rectangle.
• We can use any one of the two methods. It is convenient to use the method in (13). This is because, it has f(a+nh) instead of f(a+(n−1)h)

15. So we can write:
Area of PRSQP = $\small{\lim_{h\rightarrow 0} h\left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

• Using the result in step (4), we can replace the 'h' outside the square brackets. We get:
Area of PRSQP =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

---

Let us see a solved example:
Solved Example 23.69
Find the area bounded by the four items:
   ♦ The curve y = f(x) = x².
   ♦ The vertical line x = 2
   ♦ The vertical line x = 3
   ♦ The horizontal line y = 0 (x-axis)
Solution:
1. A rough sketch is shown in the fig.23.4 below:

Fig.23.4

• Our task is to find the area of the violet shaded area.
2. We have the formula for area:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$
• In our present case, a = 2 and b = 3
So $\small{h~=~\frac{b-a}{n}~=~\frac{3-2}{n}~=~\frac{1}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right]}$

= $\small{\lim_{n\rightarrow \infty} \frac{3-2}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$

= $\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[(a+\frac{1}{n})^2~+~(a+\frac{2}{n})^2~+~~.~.~.~+(a+\frac{n}{n})^2 \right]}$

4. Let us determine the quantity inside the square brackets:

$\small{ \left[\left(a^2 + \frac{2a}{n} + \frac{1^2}{n^2} \right)~+~\left(a^2 + \frac{4a}{n} + \frac{2^2}{n^2} \right)~+~.~.~.~+~\left(a^2 + \frac{2an}{n} + \frac{n^2}{n^2} \right) \right]}$

5. So we have to do three summations:

(i) $\small{a^2~+~a^2~+~a^2~+~.~.~.~ \text{n terms}~=~\small{n a^2~=~n(2^2)~=~4n}}$

$\small{~~~~~=~\small{n a^2~=~n(2^2)~=~4n}}$

(ii) $\small{\frac{2a}{n}\left(1~+~2~+~3~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{2a}{n}\left(\frac{n(n+1)}{2} \right)~=~\frac{a(n+1)}{2}~=~\frac{2(n+1)}{2}~=~n+1}$

(Here we use the technique of arithmetic progression)

(iii) $\small{\frac{1}{n^2}\left(1^2~+~2^2~+~3^2~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~=~\frac{1}{n^2}\left(\frac{2n^3 + 3 n^2 + n}{6} \right)~=~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

(Here we use the technique that we saw in section 9.5)

6. So the total of three summations is:

$\small{4n~+~n+1~+~\frac{n}{3}~+~\frac{1}{2}~+~\frac{1}{6n}}$

$\small{~=~\frac{2}{n}~+~\frac{19\,n}{3}-\frac{9}{2}+\frac{1}{6n}~=~\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2}}$

7. Now the limit in step (3) can be calculated:

$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{13}{6n}~+~\frac{19\,n}{3}-\frac{9}{2} \right]}$

$\small{~=~\lim_{n\rightarrow \infty}  \left[\frac{13}{6n^2}~+~\frac{19}{3}-\frac{9}{2n} \right]~=~\frac{19}{3}}$

---

Let us write the definition of definite integral. It can be written in 3 steps:
1. We have seen a method to find the area.
• Symbolically, this area is denoted as $\small{\int_a^b{\left[f(x) \right]dx}}$.
   ♦ This is the definite integral of f(x).
2. So we can write:
Area = $\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~~.~.~.~+f(a+nh) \right] ~=~\int_a^b{\left[f(x) \right]dx}}$
3. Within the square brackets, we have a sum. We are taking the limiting value of that sum. So it is a limit of sum.
• Then we can define the definite integral $\small{\int_a^b{\left[f(x) \right]dx}}$, as the limit of sum.

---

Now we will write an important point. It can be written in 3 steps:
1. In the above solved example, we calculated the definite integral of f(x) = x²
2. It is clear that the variable associated with the function f is x.
• But we applied the limit to the variable n.
3. We see that, the result will depend upon f(x), a and b. The result will not depend upon n.
• So n is called the variable of integration. It is also called the dummy variable.

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In the next section, we will see a few more solved examples.

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21.5 - Differentiability

In the previous section, we completed a discussion on continuous functions. In this section, we will see differentiability.

We have seen derivatives in class 11 (Details here). Let us recall some basic details. It can be written in 8 steps:
1. If f is a real function and c a point in it's domain, then the derivative of f at c is given by:
$f'(c) = \lim_{h\rightarrow 0} \left[\frac{f(c+h) - f(c)}{h} \right]$
2. We can find the derivative at a particular point.
• The derivative at a particular point c is denoted as f'(c).
3. We can also find the general form of the derivative.
• The general form is denoted as f'(x).
• In the general form, we can substitute c in the place of x. This will give the actual derivative at c.
4. f'(c) is the "slope of the tangent" of f at the point c.
5. f'(c) is also the "rate of change" of the output.
• For example, consider the function in which time (t) is the input and velocity (v) is the output. Then the derivative at t = 8 s will give the rate of change of velocity at the instant when the stop-watch reading is 8 s.
6. The process of finding the derivative of a function is called differentiation.
• Some text books use the following phrase:
Differentiate f(x) with respect to x.
   ♦ This means: we are being asked to find f'(x).
7. We saw three rules related to algebra of derivatives:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(u \pm v)'}    & {~=~}    &{u' \pm v'}    \\
{~\color{magenta}    2    }    &{{}}    &{(uv)'}    & {~=~}    &{u' v + u v'~\text{(Leibnitz rule or Product rule)}}    \\
{~\color{magenta}    3    }    &{{}}    &{\left(\frac{u}{v} \right)'}    & {~=~}    &{\frac{u' v \,-\, u v' }{v^2}~\text{Wherever}~v \ne 0~\text{(Quotient rule)}}    \\
\end{array}$

8. We saw some standard results also:

$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{If}}    &{f(x)}    & {~=~}    &{x^n}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{n x^{n-1}}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    2    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\sin x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{\cos x}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    3    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\cos x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{- \sin x}    \\
{~\color{magenta}    {}    }    &{{}}    &{{}}    {}    &{{}}    \\
{~\color{magenta}    4    }    &{\text{If}}    &{f(x)}    & {~=~}    &{\tan x}    \\
{~\color{magenta}    {}    }    &{\text{   Then}}    &{f'(x)}    & {~=~}    &{\sec^2 x}    \\
\end{array}$

---

Now we will derive a more convenient expression for derivative. It can be done in 5 steps:

1. Consider the definition of derivative:
$f'(x)~=~\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}$
• We obtained this expression using fig.13.31 in section 13.13.
2. The above expression can be written in another convenient form by using fig.21.14 below:

Fig.21.14

• Point P has:
   ♦ x-coordinate c
   ♦ y-coordinate f(c)
• Point Q has:
   ♦ x-coordinate x
   ♦ y-coordinate f(x)
3. So base of the triangle PQR = PR = (x-c)
Also, altitude = QR = [f(x) − f(c)]
4. Then slope of the line PQ = $\frac{f(x) - f(c)}{x - c}$
• We want PQ to be the tangent at P. So we want Q to be very close to P. That is., we want x to be very close to c. We write this as x→c.
5. When x is very close to c, the slope of PQ is given by:
$\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
• The slope in such a situation is:
the slope of tangent at P.
• Slope of tangent at P is the derivative at c.
• So we can write:
Derivative at c = $f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$

---

Now we will see the "importance of continuity" in the process of differentiation. It can be written in steps:
1. We saw that, the derivative at c is given by:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
2. The above expression can be rearranged as:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f'(c)}    & {~=~}    &{\lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{\lim_{x\rightarrow c}[f(x) - f(c)]}{\lim_{x\rightarrow c}[x - c]}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\lim_{x\rightarrow c}[f(x)]~-~\lim_{x\rightarrow c}[f(c)]}{\lim_{x\rightarrow c}[x - c]}}    \\
\end{array}$                           

3. Consider the first term in the numerator:
$\lim_{x\rightarrow c}[f(x)]$
• It shows that, limit at c must exist.
4. We know that:
If limit at c is to exist, both left side and right side limits at c must exist, and they must be the same.
5. We have seen several examples where limit at a particular point does not exist. To cite one example, see fig.21.6 in section 21.2.
• In this example, we cannot find the derivative at x = 1 because, $\lim_{x\rightarrow 1}[f(x)]$ does not exist.    
6. While discussing the topic of continuity, we saw that:
If at a point c, both left side and right side limits exist and if they are equal, then the function is continuous at c.
7. Based on this condition for continuity, we can write the condition for differentiability.
• We can write:
If we want to find the derivative of a function f at the point c, then f must be continuous at c.
8. If it is possible to find the derivative of a function f at the point c, then we say that, f is differentiable at c.
• So we can write:
For differentiability at c, "continuity at c" is essential.
9. We have seen continuous functions. Such functions are continuous at every point in the domain.
• So we can write:
If the domain of a continuous function f, is [a,b], then f is differentiable at every point in [a,b]
• As in the case of continuity,
   ♦ At the point a, we need to consider only the right side limit.
   ♦ At the point b, we need to consider only the left side limit.

---

Now we will see a theorem which gives the relation between continuity and differentiability.
• The theorem states that:
If a function f is differentiable at a point c, then the function is continuous at c.

The proof can be written in 3 steps:
1. Consider the expression for derivative that we obtained above:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$ 
2. Now consider the case when x ≠ c.
• We can write f(x) - f(c) as:
$f(x) - f(c) = {\frac{f(x) - f(c)}{x-c}}.(x-c)$
• Taking limits on both sides, we get:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c} \left[f(x) - f(c) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}}.(x-c) \right]}    \\
{~\color{magenta}    2    }    &{\implies}    &{\lim_{x\rightarrow c} \left[f(x) \right] - \lim_{x\rightarrow c} \left[f(c) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[{\frac{f(x) - f(c)}{x-c}} \right].\lim_{x\rightarrow c} \left[(x-c) \right]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{f'(c) . 0}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{\implies}    &{\lim_{x\rightarrow c} \left[f(x) \right]}    & {~=~}    &{\lim_{x\rightarrow c} \left[f(c) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\lim_{x\rightarrow c} f(x)}    & {~=~}    &{f(c)}    \\
\end{array}$                           

3. We want to prove that, f is continuous at c. Recall that, we need to check two conditions to prove continuity.
(i) Limit exists at c.
(ii) $\lim_{x\rightarrow c} f(x) = f(c)$
• In the theorem, it is given that, f is differentiable at c. So it is obvious that, limit at c exists. Thus condition (i) is satisfied.
• From (3) we have: $\lim_{x\rightarrow c} f(x) = f(c)$.
Thus condition (ii) is also satisfied.
◼ Hence it is proved that, f is continuous at c.  

◼ As a corollary of the above theorem, we can write:
Every differentiable function is continuous.
(Corollary of a theorem, is a fact, which results directly from that theorem. The dictionary meaning can be seen here)

The converse of the above corollary is not true. That is., every continuous function is not differentiable. Let us see an example. It can be written in steps:
1. Consider the function f(x) = |x|.
• We know that, it is a continuous function. We want to prove that, it is not a differentiable function.
2. Let us try to find the derivative of this function at x = 0
3. We have the general equation:
$f'(c) = \lim_{x\rightarrow c} \left[\frac{f(x) - f(c)}{x - c} \right]$
• So the derivative at x = 0 is given by:
$f'(0) = \lim_{x\rightarrow 0} \left[\frac{|x| - f(0)}{x - 0} \right]$
4. Let us write the left side limit at x = 0:
$\lim_{x\rightarrow 0^{-}} \left[\frac{|x| - f(0)}{x - 0} \right]$
• As x approaches zero from the left side, the input will be a −ve value. Let it be −h.
• So the left side limit will be:
$\frac{|-h| - |0|}{(-h) - 0} ~=~\frac{h}{(-h)}~=~-1$ 
5. Let us write the right side limit at x = 0:
$\lim_{x\rightarrow 0^{+}} \left[\frac{|x| - f(0)}{x - 0} \right]$
• As x approaches zero from the right side, the input will be a +ve value. Let it be h.
• So the right side limit will be:
$\frac{|h| - |0|}{h - 0}~=~\frac{h}{h}~=~1$   
6. From (4) and (5), we see that:
Left side and right side limits are not the same. So the required limit does not exist at x = 0.
• Therefore, the derivative at x = 0 does not exist.
7. Based on this example, we can write:
All continuous functions are not differentiable.

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In the next section, we will see derivatives of composite functions.

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21.4 - Solved Examples on Algebra of Continuous Functions

In the previous section, we saw algebra of continuous functions. In this section, we will see some solved examples.

Solved Example 21.16
Prove that every rational function is continuous.
Solution:
1. Every rational function will be of the form: $f(x) = \frac{p(x)}{q(x)}$
    ♦ p(x) and q(x) are polynomial functions.
    ♦ Also q(x) should not be zero.
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3. In our present case, both numerator and denominator are polynomial functions. We have seen that all polynomial functions are continuous.
• So we can write: all rational functions are continuous.
4. We must consider the domain of a given rational function. That domain will not contain any real number which makes q(x) equal to zero.
• So for all values in the domain, the given rational function will be indeed continuous.

Solved Example 21.17
Discuss the continuity of the sine function.
Solution:
1. Fig.21.11 below shows the graph of the sine function f(x) = sin x.

Fig.21.11

2. An arbitrary point c is marked on the graph.
We want to find $\lim_{x\rightarrow c^{-}} f(x)$  and $\lim_{x\rightarrow c^{+}} f(x)$ 

3. First we will find $\lim_{x\rightarrow c^{-}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the left of c. We can write it as x = c−h
(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c^{-}} f(x)}    & {~=~}    &{\lim_{h\rightarrow 0} \sin (c-h)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h - \cos c \, \sin h]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h] ~-~\lim_{h\rightarrow 0} [\cos c \, \sin h]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \lim_{h\rightarrow 0} [\cos h] ~-~\cos c \, \lim_{h\rightarrow 0} [ \sin h]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \cos 0 ~-~\cos c \,  \sin 0}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, (1) ~-~\cos c \, (0)}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\sin c}    \\
\end{array}$

◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a−b) = sin a cos b − cos a sin b
• 5 (magenta color):
    ♦ From the graph of the cosine function, it is clear that, when the input value approaches zero from left or right, the limiting cosine value is 1.
    ♦ From the graph of the sine function, it is clear that, when the input value approaches zero from left or right, the limiting sine value is 0.

4. Next we will find $\lim_{x\rightarrow c^{+}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the right of c. We can write it as x = c+h. This is shown in the graph below:

Fig.21.12

(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c^{+}} f(x)}    & {~=~}    &{\lim_{h\rightarrow 0} \sin (c+h)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h + \cos c \, \sin h]}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\lim_{h\rightarrow 0} [\sin c \, \cos h] ~+~\lim_{h\rightarrow 0} [\cos c \, \sin h]}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \lim_{h\rightarrow 0} [\cos h] ~+~\cos c \, \lim_{h\rightarrow 0} [ \sin h]}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, \cos 0 ~+~\cos c \,  \sin 0}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{\sin c \, (1) ~+~\cos c \, (0)}    \\
{~\color{magenta}    7    }    &{{}}    &{{}}    & {~=~}    &{\sin c}    \\
\end{array}$

◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a+b) = sin a cos b + cos a sin b

5. We see that, left side limit is equal to the right side limit. That means, limit exists.

6. We know that, the value of the function at x = c is:
f(c) = sin c

7. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = sin x is a continuous function.

---

In a similar way, we can prove that, the cosine function is also a continuous function. Steps are shown briefly below:

1. Left side limit can be calculated as:

◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a−b) = cos a cos b + sin a sin b

2. Right side limit can be calculated as:

◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a+b) = cos a cos b − sin a sin b

3. We know that, the value of the function at x = c is:
f(c) = cos c

4. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = cos x is a continuous function.

---

Solved Example 21.18
Discuss the continuity of the function f(x) = tan x.
Solution:
1. f(x) = tan x can be written as: $\rm{f(x) = \frac{\sin x}{\cos x}}$
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3. In our present case, we just saw in the above solved example that, both sine and cosine functions are continuous functions.
• So we can write: f(x) = tan x is a continuous function.
4. We must consider the domain of this function. That domain will not contain any real number which makes 'cos x' equal to zero.
• So for all values in the domain, f(x) = tan x will be indeed continuous.
5. More details about the domain can be obtained from the graph below. It is the graph of f(x) = tan x.

Fig.21.13

• We see that:
When the input x value approaches $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the f(x) value approaches infinity. This is because, the denominator becomes smaller and smaller and approaches zero. When the input x values are exact $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the denominator becomes zero. Division by zero will give a number which is not defined. So we must avoid these input values in the domain. In short, the domain should not contain those x values which are given by $\rm{(2n+1){\frac{\pi}{2}}}$, where n is any integer, +ve or -ve.

---

Now we will see continuity of composite functions. It can be written in 4 steps:
1. Consider the composite function (f∘g)(x). It can also be written as f(g(x)).
• We see that, the output of g is being used as the input for f.
2. Consider an arbitrary point c. When the input for g is c, the output will be g(c).
• Then the input for f will be g(c)
3. When we consider the two functions f and g together, we see that, two inputs are being made:
    ♦ c is the input for g
    ♦ g(c) is the input for f
4. Now we can write about the continuity of (f∘g).
    ♦ Suppose that, g is continuous at c.
    ♦ Also suppose that, f is continuous at g(c)
• Then we can write: (f∘g) is continuous at c.
We will see the proof in higher classes.

Solved Example 21.19
Show that the function defined by f(x) = sin (x²) is a continuous function.
Solution:
1. The given function is: f(x) = sin (x²)
2. We can write it as the composite of two functions: f(x) = g(h(x))
    ♦ Where h(x) = x² and g(h(x)) = sin (h(x))
3. Consider h(x) = x²
• At any arbitrary point c, we know that, h will be continuous.
• The output of h at c is h(c) = c².
4. So the input for g at c is c².
• c² is a real number. We know that sin x is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = sin (x²) is a continuous function.

Solved Example 21.20
Show that the function defined by f(x) = |1− x + |x|| is a continuous function.
Solution:
1. The given function is: f(x) = |1 − x + |x||
2. We can write it as the composite of two functions: f(x) = g(h(x))
    ♦ Where h(x) = 1 − x + |x| and g(h(x)) = |h(x)|
3. Consider h(x) = 1 − x + |x|
• This is the sum of two continuous functions:
(1−x) and |x|
    ♦ (1−x) is a polynomial function. So it is continuous.
    ♦ |x| is the modulus function. It is also continuous.
• Thus h, which is the sum of two continuous functions, will be continuous.
• In other words, at any arbitrary point c, h will be continuous.
• The output of h at c is h(c) = 1 − c + |c|.
4. So the input for g at c is (1 − c + |c|).
• (1 − c + |c|) is a real number. We know that g(x) = |x| is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = |1 − x + |x|| is a continuous function.

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The link to a few more solved examples is given below:

Exercise 21.1 (Parts 1 and 2)

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In the next section, we will see differentiability.

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21.3 - Algebra of Continuous Functions

In the previous section, we saw how to check whether a given function is a continuous function or not. In this section, we will see algebra of continuous functions.

• In class 11, we have seen algebra of limits. (Details here). Now in the previous few sections, we saw that, limits are the deciding factors for continuous functions. So naturally, we can think about algebra of continuous functions.
• Suppose that, f and g are two real functions.
• Also suppose that:
    ♦ f is continuous at c.
    ♦ g is continuous at c.
• Then we can write four results:
    ♦ f+g is continuous at x = c
    ♦ f−g is continuous at x = c
    ♦ f.g is continuous at x = c
    ♦ $\frac{f}{g}$ is continuous at x = c (provided g(c) ≠ 0)

---

• We will write the proof for the first result. It can be written in 4 steps:
1. We want to prove that (f+g) is continuous at c.
2. Let us apply the first condition.
That is: $\lim_{x\rightarrow c} (f+g)(x)$ must exist.
• Let us check whether this is true.

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\lim_{x\rightarrow c} (f+g)(x)}    & {~=~}    &{\lim_{x\rightarrow c} [f(x) + g(x)]}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\lim_{x\rightarrow c} f(x)~+~\lim_{x\rightarrow c} g(x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{f(c) ~+~ g(c)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{(f+g)(c)}    \\
\end{array}$

◼ Remarks:
• 1 (magenta color). Here we use the fact about addition of functions.
(f+g)(x) = f(x) + g(x)
2 (magenta color). Here we use the algebra of limits that we saw in class 11.
3 (magenta color). We get this result because, it is given that, f and g are continuous at c.
4 (magenta color). Here we again use the fact about addition of functions.
(f+g)(x) = f(x) + g(x)

So $\lim_{x\rightarrow c} (f+g)(x)$ exists.

3. Let us apply the second condition:
$\lim_{x\rightarrow c} (f+g)(x)$ must be equal to (f+g)(c)
This is already proved in (2) above.

4. Since both conditions are satisfied, (f+g)(x) is continuous at c.

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The proofs for the remaining three results can be written in a similar way. The reader is advised to write those proofs in his/her own notebooks.

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Now we will see two special cases.

Case 1
This can be written in 4 steps.
1. Consider result 3:
If both f and g are continuous at c, then f.g is also continuous at c.
2. Suppose that, one of the two functions, say f, is a constant function. Then we can write: f(x) = λ, where λ is a real number.
(Recall that, any constant function is a continuous function)
3. Then we can write (λ.g)(x) is a continuous function.
• This is same as: λ[g(x)] is a continuous function.
(We saw this result in class 11, in the topic of multiplication of functions)
4. From this, we get an interesting result:
If λ = −1, then −g(x) is a continuous function.
• So, if g is a continuous function, then -g is also a continuous function.

Case 2
This can be written in 4 steps.
1. Consider result 4:
If both f and g are continuous at c, then $\frac{f}{g}$ is also continuous at c.
2. Suppose that, f is a constant function. Then we can write: f(x) = λ, where λ is a real number.
(Recall that, any constant function is a continuous function)
3. Then we can write $\left( \frac{\lambda}{g} \right) (x)$ is a continuous function.
• This is same as: $\lambda \left[\left( \frac{1}{g} \right) (x) \right]$ is a continuous function.
4. From this, we get an interesting result:
If λ = 1, then $\left( \frac{1}{g} \right) (x)$ is a continuous function.
• So, if g is a continuous function, then $\frac{1}{g}$ is also a continuous function.

This case can be used to solve many problems.

---

We have seen the algebra of limits. Using those results, we can prove that, any polynomial function is continuous. It can be written in steps:

1. We want to prove that
$f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ~.~.~.~+ a_n x^n$
is continuous.
• We can use mathematical induction.

2. First, we check whether the function is continuous when n = 1.
• That is., we want to know whether $f(x) = a_0 + a_1 x$ is continuous.
• It is indeed continuous because, it is the sum of two continuous functions: $f_1 (x) = a_0 $ and $f_2 (x) = a_1 x$

3. Next, we assume that, the function is continuous when n = k.
This can be written as follows:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ~.~.~.~+ a_k x^k}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{a_0 + x(a_1 + a_2 x^1 + a_3 x^2 + ~.~.~.~+ a_{k} x^{k-1})}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{f_1 (x) + x[f_3 (x)]}    \\
\end{array}$                           

• So f(x) is the sum of two functions:
(i) $f_1 (x) = a_0 $
(ii) $x[f_3 (x)] = x(a_1 + a_2 x^1 + a_3 x^2 + ~.~.~.~+ a_{k} x^{k-1}) $

• We assume that, this f(x) is continuous. If f(x) is continuous, then it's components will also be continuous. That is.,
    ♦ $f_1 (x)$ is continuous.
    ♦ $x[f_3 (x)]$ is continuous.
• If $x[f_3 (x)]$ is continuous, then $f_3 (x)$ will be continuous.
• So, by assuming that the function is continuous when n = k, we get an important result:
$f_3 (x)$ is continuous.

4. Next, we check whether the function is continuous when n = k+1.
This can be written as follows:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{f(x)}    & {~=~}    &{a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ~.~.~.~+ a_k x^k + a_{k+1} x^{k+1}}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{a_0 + x(a_1 + a_2 x^1 + a_3 x^2 + ~.~.~.~+ a_{k} x^{k-1} + a_{k+1} x^k)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{a_0 + x(a_1 + a_2 x^1 + a_3 x^2 + ~.~.~.~+ a_{k} x^{k-1}) + x \, a_{k+1} x^k}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{f_1 (x) + x[f_3 (x)] + a_{k+1} x^{k+1}}    \\
\end{array}$                           

• So this f(x) is the sum of three components.
(i) We already know that, $f_1 (x)$ is continuous.
(ii) We already know that, $x[f_3 (x)]$ is continuous.
(iii) The third component is also continuous because, it is the product of $a_{k+1}$ and x taken (k+1) times.

• So f(x) is continuous.  
    ♦ That means, the function is continuous when k = 1.
    ♦ Also, if the function is continuous for n = k, it will be continuous for n = (k+1)
    ♦ So we prove the continuity by mathematical induction.

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In the next section, we will see some solved examples.

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21.1 - Continuous Functions

In the previous section, we saw the two conditions which can be used to check continuity at any given point.

• In the solved examples that we saw in the previous section, we were checking the continuity at specified points like x = 0, x = 1 etc.,
• Now let us see some examples where the given functions are continuous at every point.

Solved example 21.5
Check the points where the constant function f(x) = k is continuous.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: k
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = k
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number.

Solved example 21.6
Prove that the identity function on real numbers given by f(x) = x is continuous at every real number.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number.

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• In the above two solved examples 21.5 and 21.6, we see that, the given functions are continuous at every point. We say that, such functions are continuous functions. For such functions, there is no need to check for continuity at various individual points.
• We can write the definition for continuous functions:
A real function f is said to be a continuous function, if it is continuous at every point in the domain of f.

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Now we will see a special case. It can be written in steps:
1. In the solved examples 21.5 and 21.6, the domain is (−∞,∞). The functions are continuous between −∞ and ∞.
2. Let us see the case where domain is restricted. In fig.21.2 below, we see that, the domain is [a,b].

Fig.21.2

3. In the above fig., If the function f is to be a continuous function, then:
   ♦ f must be continuous at A.
   ♦ f must be continuous at B.
   ♦ f must be continuous at all the infinite points in between A and B.
4. If f is to be continuous at A, $\lim_{x\rightarrow a^{-}} f(x)$ should be equal to $\lim_{x\rightarrow a^{+}} f(x)$.
• But there is no way to find $\lim_{x\rightarrow a^{-}} f(x)$. This is because, we do not know how x approaches 'a' from the left.
• So there is no need to consider $\lim_{x\rightarrow a^{-}} f(x)$.
• To check the continuity at A, we need to check only one condition: $\lim_{x\rightarrow a^{+}} f(x) ~=~f(a)$ 
5. Similarly, if f is to be continuous at B, $\lim_{x\rightarrow b^{-}} f(x)$ should be equal to $\lim_{x\rightarrow b^{+}} f(x)$.
• But there is no way to find $\lim_{x\rightarrow b^{+}} f(x)$. This is because, we do not know how x approaches 'b' from the right.
• So there is no need to consider $\lim_{x\rightarrow b^{+}} f(x)$.
• To check the continuity at B, we need to check only one condition: $\lim_{x\rightarrow b^{-}} f(x) ~=~f(b)$
6. Now consider the case when the domain of f is a singleton. That is., the domain is a set which contains only one point.
• Then the graph will be a point. We cannot check left side limit or right side limit.
• In such a situation, we say that, f is a continuous function.

Let us see some solved examples:

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Solved example 21.7
Is the function defined by f(x) = |x|, a continuous function?
Solution:
• Fig.21.3 below shows the graph of the function f(x) = |x|.

Fig.21.3

Case 1: continuity of the left segment.
1. Consider any arbitrary point P on the left segment. Let us check whether the function is continuous at P.
2. Applying condition (i):
• Limiting value of f(x) at x = -p is: p
• It is clear that, $\lim_{x\rightarrow -p} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(-p) = |-p| = p
• We see that: $\lim_{x\rightarrow -p} f(x)~=~f(-p)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at P.
5. P was taken as an arbitrary point. So '-p' can be any real number less than zero.
• Therefore we can write:
The given function is continuous at every real number less than zero.

Case 2: continuity of the right segment.
1. Consider any arbitrary point Q on the right segment. Let us check whether the function is continuous at Q.
2. Applying condition (i):
• Limiting value of f(x) at x = q is: q
• It is clear that, $\lim_{x\rightarrow q} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(q) = |q| = q
• We see that: $\lim_{x\rightarrow q} f(x)~=~f(q)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at Q.
5. Q was taken as an arbitrary point. So q can be any real number greater than or equal to zero.
• Therefore we can write:
The given function is continuous at every real number greater than or equal to zero.

◼ Based on the results from the two cases. we can write:
The given function is continuous at all points.

Solved example 21.8
Discuss the continuity of the function defined by:
f(x) = x³ + x² − 1.
Solution:
1. Consider any arbitrary point c. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c³ + c² − 1
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c³ + c² − 1
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point. So it can be any real number.
• Therefore we can write:
The given function is continuous at every real number. Therefore, it is a continuous function.

Solved example 21.9
Is the function defined by $f(x) = \frac{1}{x},~x \ne 0$, a continuous function?
Solution:
• Fig.21.4 below shows the graph of the function $f(x) = \frac{1}{x}$.

Fig.21.4

Case 1: continuity of the left segment.
1. Consider any arbitrary point P on the left segment. Let us check whether the function is continuous at P. (Note that, it is impossible to consider any point whose x coordinate is zero. It is specially mentioned in the question)
2. Applying condition (i):
• Limiting value of f(x) at x = -p is: $-{\frac{1}{p}}$
• It is clear that, $\lim_{x\rightarrow -p} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(-p) = $\frac{1}{(-p)}~=~-{\frac{1}{p}}$
• We see that: $\lim_{x\rightarrow -p} f(x)~=~f(-p)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at P.
5. P was taken as an arbitrary point. So '-p' can be any real number less than zero.
• Therefore we can write:
The given function is continuous at every real number less than zero. However, the real number 'zero' can not be considered.

Case 2: continuity of the right segment.
1. Consider any arbitrary point Q on the right segment. Let us check whether the function is continuous at Q.
2. Applying condition (i):
• Limiting value of f(x) at x = q is: $\frac{1}{q}$
• It is clear that, $\lim_{x\rightarrow q} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(q) = $\frac{1}{q}$
• We see that: $\lim_{x\rightarrow q} f(x)~=~f(q)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at Q.
5. Q was taken as an arbitrary point. So 'q' can be any real number greater than zero.
• Therefore we can write:
The given function is continuous at every real number greater than zero. However, the real number 'zero' can not be considered.

◼ Based on the results from the two cases. we can write:
The given function is continuous at all real numbers other than zero.

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Based on the above solved example 21.9, we can discuss about the concept of infinity.
The graph is shown again in fig.21.5 below:

Fig.21.5

Case 1: The right segment.
This can be written in 5 steps:
1. Two points Q₁ and Q₂ are marked on the right segment.
   ♦ Q₂ is closer (horizontally) to zero than Q₁.
   ♦ So q₂ will be smaller than the q₁.
2. The input x values are given in the denominator.
• So smaller the denominator, larger will be the value of the function f(x).
• So f(q₂) will be larger than f(q₁).
• It is clear that, as x approaches zero from the right, f(x) will become larger and larger.
3. When x approaches zero from the right, it take values like 0.1, 0.001, 0.0001, etc.,     
• Let us find f(x) in such cases:
   ♦ When x = 0.1, f(x) = $\frac{1}{0.1} = 10$
   ♦ When x = 0.01, f(x) = $\frac{1}{0.01} = 100$
   ♦ When x = 0.001, f(x) = $\frac{1}{0.001} = 1000$
   ♦ When x = 0.0001, f(x) = $\frac{1}{0.0001} = 10000$
4. In the input x, we can give a million zeros after the decimal point. The resulting f(x) will be correspondingly large.
• We can make f(x) larger than the largest known number. All we need to do is to give the required number of zeros after the decimal point.
• "Larger than the largest known number" is denoted by the symbol +∞. It is read as plus infinity.
• +∞ is a concept. It is not a real number. We cannot do familiar calculations with +∞.
• For example:
   ♦ (+∞ + 4) gives +∞
   ♦ (+∞ ÷ 9) gives +∞
• We assume that:
+∞ is at the right end of the x-axis and at the top end of the y-axis.
5. So when $f(x) = \frac{1}{x}$, we can write: $\lim_{x\rightarrow 0^{+}} f(x)~=~+ \infty$

Case 2: The left segment.
This can be written in 5 steps:
1. Two points P₁ and P₂ are marked on the left segment.
   ♦ P₂ is closer (horizontally) to zero than P₁.
   ♦ So p₂ will be smaller (numerically) than the p₁.
2. The input x values are given in the denominator.
• So smaller the denominator, larger will be the value of the function f(x).
• So f(p₂) will be larger (numerically) than f(p₁).
• It is clear that, as x approaches zero, f(x) will become larger and larger numerically.
3. But on the left segment, the x values are −ve.
• If a "numerically larger value" is −ve, it is smaller in reality.
• So we can write:
As x approaches zero from the left, f(x) will become smaller and smaller.
4. When x approaches zero, it take values like −0.1, −0.001, −0.0001, etc.,     
• Let us find f(x) in such cases:
   ♦ When x = −0.1, f(x) = $\frac{1}{-0.1} = -10$
   ♦ When x = −0.01, f(x) = $\frac{1}{-0.01} = -100$
   ♦ When x = −0.001, f(x) = $\frac{1}{-0.001} = -1000$
   ♦ When x = −0.0001, f(x) = $\frac{1}{-0.0001} = -10000$
5. In the input x, we can give a million zeros after the decimal point. The resulting f(x) will be correspondingly small.
• We can make f(x) smaller than the smallest known number. All we need to do is to give the required number of zeros after the decimal point.
• "Smaller than the smallest known number" is denoted by the symbol −∞. It is read as minus infinity.
• −∞ is a concept. It is not a real number. We cannot do familiar calculations with −∞.
• For example:
   ♦ (−∞ + 4) gives −∞
   ♦ (−∞ ÷ 9) gives −∞
• We assume that:
−∞ is at the left end of the x-axis and at the bottom end of the y-axis.
6. So when $f(x) = \frac{1}{x}$, we can write: $\lim_{x\rightarrow 0^{−}} f(x)~=~− \infty$

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In the next section, we will see a few more solved examples.

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