Chapter 13.6 - Examples on Limits of Rational Functions

In the previous section, we saw limits of polynomial functions and rational functions. In this section, we will see some solved examples.

Solved example 13.1
(i) Find the limit of f(x) = x³ - x² + 1 when x approaches 1
(ii) Find the limit of f(x) = x(x+1) when x approaches 3
(iii) Find the limit of
f(x) = 1 + x + x² + x³ + . . . + x¹⁰ when x approaches -1
Solution:
Part (i):
The given function is a polynomial function. For a polynomial function, the limiting value is same as the value of the function at the required point. So we can write:
$$\lim_{x\rightarrow 1} f(x) = f(1) = (1^3 - 1^2 + 1) = (1 - 1 + 1) = 1$$

Part (ii):
The given function is a polynomial function. For a polynomial function, the limiting value is same as the value of the function at the required point. So we can write:
$$\lim_{x\rightarrow 3} f(x) = f(3) = 3(3 + 1) = 3 × 4 = 12$$

Part (iii):
The given function is a polynomial function. For a polynomial function, the limiting value is same as the value of the function at the required point. So we can write:
$$\lim_{x\rightarrow -1} f(x) = f(1) = \left[1 + (- 1) + (-1)^2 + (-1)^3 ~+~.~.~.~+~(-1)^{10} \right] = [1 + 1 -1 +1~+~.~.~.~+~1] = 1$$
Note: The sum inside square brackets can be calculated using the formula: $S_n = \frac{a \left(r^n - 1 \right)}{r-1}$
• We saw this in a previous chapter (details here).
• Here, a = 1, r = -1 and n = 11

Solved example 13.2
(i) Find the limit of $f(x) = \frac{x^2 + 1}{x + 100}$ when x approaches 1
(ii) Find the limit of $f(x) = \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$ when x approaches 2
(iii) Find the limit of $f(x) = \frac{x^2 - 4}{x^3 - 4x^2 + 4x}$ when x approaches 2
(iv) Find the limit of $f(x) = \frac{x^3 - 2x^2}{x^2 - 5x + 6}$ when x approaches 2
(v) Find the limit of
$f(x) = \left[\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}\right]$ when x approaches 1
Solution:
Part (i):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\

{}&{}
& {~=~}& {\frac{1^2 + 1}{1 + 100}}
&{} \\

{}&{}
& {~=~}& {\frac{2}{101}}
&{} \\

\end{array}$

Part (ii):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2^3 - 4 × 2^2 + 4 × 2}{2^2 - 4}}
&{} \\

{}&{}
& {~=~}& {\frac{8 - 16 + 8}{4 - 4}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\

\end{array}$

4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:

$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^3 - 4x^2 + 4x}{x^2 - 4}}
&{} \\

{}&{}
& {~=~}& {\frac{x(x^2 - 4x + 4)}{(x+2)(x-2)}}
&{} \\

{}&{}
& {~=~}& {\frac{x(x-2)^2}{(x+2)(x-2)}}
&{} \\

{}&{}
& {~=~}& {\frac{x(x-2)}{(x+2)}}
&{} \\

\end{array}$

5. Now we can find the limit:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2(2-2)}{(2+2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2 × 0}{4}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{4}}
&{} \\

{}&{}
& {~=~}& {0}
&{} \\

\end{array}$

6. The graph in fig.13.18 below shows that:
When x approaches 2, the value of f(x) indeed approaches zero.

Fig.13.18

7. Note:
This can be written in 3 steps:
(i) In step (4), we canceled (x-2) from both numerator and denominator.
(ii) If we are doing the actual evaluation of f(x) at x = 2, then such a cancellation will not be possible. This is because, if it is an actual evaluation, x will be exactly ‘2’. Then (x-2) will be zero. Division by zero will give a number which does not exist.
(iii) We did the cancellation because, in our present case, it is not an actual evaluation. It is the evaluation of the limiting value. While evaluating the limiting value, x is not exactly equal to 2. So (x-2) will not be zero.

Part (iii):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2^2 - 4}{2^3 - 4 × 2^2 + 4 × 2}}
&{} \\

{}&{}
& {~=~}& {\frac{4 - 4}{8 - 16 + 8}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\

\end{array}$

4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:

$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^2 - 4}{x^3 - 4x^2 + 4x}}
&{} \\

{}&{}
& {~=~}& {\frac{(x+2)(x-2)}{x(x^2 - 4x + 4)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x+2)(x-2)}{x(x-2)^2}}
&{} \\

{}&{}
& {~=~}& {\frac{(x+2)}{x(x-2)}}
&{} \\

\end{array}$

5. Now we can find the limit:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{(2+2)}{2(2-2)}}
&{} \\

{}&{}
& {~=~}& {\frac{4}{2 × 0}}
&{} \\

{}&{}
& {~=~}& {\frac{4}{0}}
&{} \\

\end{array}$

6. We can write:
The limit of the given f(x) is not defined when x approaches 2

7. This is evident from the graph in fig.13.19 below:

Fig.13.19

(i) The graph has a particular shape when x is less than zero
(ii) The graph has another shape when x lies between zero and 2
(iii) The graph has yet another shape when x is greater than 2
(iv) We see that:
• When x approaches 2 from the left, f(x) approaches -∞.  
• When x approaches 2 from the right, f(x) approaches +∞.
(v) So for this function, the limit when x approaches 2 is not defined.  

Part (iv):
1. The given f(x) is already written in the form $\frac{g(x)}{h(x)}$
2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2^3 - 2 × 2^2}{2^2 - 5 × 2 + 6}}
&{} \\

{}&{}
& {~=~}& {\frac{8 - 8}{4 - 10 + 6}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\

\end{array}$

4. Both numerator and denominator are zeroes. So we will factorize the given function. We can write:

$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^3 - 2x^2}{x^2 - 5x + 6}}
&{} \\

{}&{}
& {~=~}& {\frac{x^2(x - 2)}{(x-3)(x-2)}}
&{} \\

{}&{}
& {~=~}& {\frac{x^2}{(x-3)}}
&{} \\

\end{array}$

5. Now we can find the limit:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 2} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 2} g(x)}{\lim_{x\rightarrow 2} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(2)}{h(2)}}
&{} \\

{}&{}
& {~=~}& {\frac{2^2}{2-3}}
&{} \\

{}&{}
& {~=~}& {\frac{4}{-1}}
&{} \\

{}&{}
& {~=~}& {-4}
&{} \\

\end{array}$

Part (v):
$f(x) = \left[\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}\right]$ when x approaches 1
 

1. First we will write the given function in the form: $f(x) = \frac{g(x)}{h(x)}$

It can be done as follows:
$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x - 2}{x^2 - x}~-~\frac{1}{x^3 - 3x^2 + 2x}}
&{} \\

{}&{}
& {~=~}& {\frac{x - 2}{x(x - 1)}~-~\frac{1}{x(x^2 - 3x + 2)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x - 2)(x^2 - 3x + 2)~-~(x-1)}{x(x - 1)(x^2 - 3x + 2)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x - 2)(x - 2)(x-1)~-~(x-1)}{x(x - 1)(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x-1) \left[(x - 2)(x - 2)~-~1 \right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x-1) \left[(x - 2)^2~-~1 \right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x-1) \left[x^2 - 4x + 4 - 1\right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x-1) \left[x^2 - 4x + 3\right]}{x(x - 1)(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{x^2 - 4x + 3}{x(x - 2)(x-1)}}
&{} \\

\end{array}$ 

2. We see that, both g(x) and f(x) are polynomial functions.
3. So we can write:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\

{}&{}
& {~=~}& {\frac{1^2 - 4 × 1 + 3}{1(1-2)(1-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{1 - 4 + 3}{1 × -1 × 0}}
&{} \\

{}&{}
& {~=~}& {\frac{0}{0}}
&{} \\

\end{array}$

4. Both numerator and denominator are zeroes. So we will factorize f(x). We can write:

$\begin{array}{ll}
{}&{f(x)}
& {~=~}& {\frac{x^2 - 4x + 3}{x(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{(x - 3)(x-1)}{x(x - 2)(x-1)}}
&{} \\

{}&{}
& {~=~}& {\frac{x-3}{x(x-2)}}
&{} \\

\end{array}$

5. Now we can find the limit:

$\begin{array}{ll}
{}&{\lim_{x\rightarrow 1} f(x)}
& {~=~}& {\frac{\lim_{x\rightarrow 1} g(x)}{\lim_{x\rightarrow 1} h(x)}}
&{} \\

{}&{}
& {~=~}& {\frac{g(1)}{h(1)}}
&{} \\

{}&{}
& {~=~}& {\frac{1-3}{1(1-2)}}
&{} \\

{}&{}
& {~=~}& {\frac{-2}{-1}}
&{} \\

{}&{}
& {~=~}& {2}
&{} \\

\end{array}$

6. Note:
This can be written in 3 steps:
(i) In step (4), we canceled (x-1) from both numerator and denominator.
(ii) If we are doing the actual evaluation of f(x) at x = 1, then such a cancellation will not be possible. This is because, if it is an actual evaluation, x will be exactly ‘1’. Then (x-1) will be zero. Division by zero will give a number which does not exist.
(iii) We did the cancellation because, in our present case, it is not an actual evaluation. It is the evaluation of the limiting value. While evaluating the limiting value, x is not exactly equal to 1. So (x-1) will not be zero.

---

In the next section, we will see the second theorem. 

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Chapter 13.4 - Examples For Multiplication And Division

In the previous section, we saw the theorem related to algebra of limits. We also saw examples of addition and subtraction of individual limits. In this section, we will see examples for multiplication and division.

Example for multiplication:
• This can be written in 5 steps:
1. Let us write the functions:
(i) Let the first function be: f(x) = x + 4
• This is plotted in green color in fig.13.15 below.
(ii) Let the second function be: g(x) = x² + 1
• This is plotted in yellow color in fig.13.15 below.
(iii) So the product of the two functions will be:
f(x) . g(x) = (x+4)(x²+1) = (x³ + x + 4x² + 4) = x³ + 4x² + x + 4
• This is plotted in red color in fig.13.15 below.

Fig.13.15

2. Let us write the limit of f(x) as x approaches -2
• We get: $\lim_{x\rightarrow -2} f(x) = (-2 + 4) = 2$
• Indeed, we have a green filled circle at (-2,2)
3. Let us write the limit of g(x) as x approaches -2
• We get: $\lim_{x\rightarrow -2} g(x) = [(-2)^2 + 1] = (4+1) = 5$
• Indeed, we have a yellow filled circle at (-2,5)   
4. Let us write the limit of [f(x) . g(x)] as x approaches -2
• We get: $\lim_{x\rightarrow -2} [f(x) . g(x)] = [(-2)^3 + 4(-2)^2  -2 + 4] = (-8+16-2+4) = 10$
• Indeed, we have a red filled circle at (-2,10)
5. Comparing the above steps (2), (3) and (4), we can write:
$$\lim_{x\rightarrow -2} [f(x) . g(x)]= 10 = \lim_{x\rightarrow -2} f(x) \, . \, \lim_{x\rightarrow -2} g(x) = 2 \, . \, 5$$

Example for the special case of multiplication:
• This can be written in 5 steps:
1. Let us write the functions:
(i) Let the first function be: f(x) = x² + 1
• This is plotted in green color in fig.13.16 below.
(ii) Let the second function be: g(x) = 3
• This is plotted in yellow color in fig.13.16 below.
(iii) So the product of the two functions will be:
f(x) . g(x) = (x²+1).3 = 3x² + 3
• This is plotted in red color in fig.13.16 below.

Fig.13.16

2. Let us write the limit of f(x) as x approaches 1
• We get: $\lim_{x\rightarrow 1} f(x) = (1^2 + 1) = (1+1) =2$
• Indeed, we have a green filled circle at (1,2)
3. Let us write the limit of g(x) as x approaches 1
• Here g(x) is a constant function. The constant $\lambda$ = 3.
   ♦ So whatever be the value of x, the limiting value will be '3'.
• We get: $\lim_{x\rightarrow 1} g(x) = 3$
• The point (1,3) is marked with a yellow filled circle.
4. Let us write the limit of [f(x) . g(x)] as x approaches -2
• We get: $\lim_{x\rightarrow 1} [f(x) . g(x)] = [3(1)^2 + 3] = (3+3) = 6$
• Indeed, we have a red filled circle at (1,6)
5. Comparing the above steps (2), (3) and (4), we can write:
$$\lim_{x\rightarrow 1} [f(x) . g(x)]= 6 = \lim_{x\rightarrow 1} f(x) \, . \, \lim_{x\rightarrow 1} g(x) = 2 \, . \, 3 = \lambda \, . \, \lim_{x\rightarrow 1} f(x) = 3 \, . \, 2$$
• That means:
   ♦ We need not multiply the two functions.
   ♦ We just need to take the limit of f(x) and then multiply it by $\lambda$

---

Example for division:
• This can be written in 5 steps:
1. Let us write the functions:
(i) Let the first function be: f(x) = x² + 1
• This is plotted in green color in fig.13.17 below.
(ii) Let the second function be: g(x) = x + 4
• This is plotted in yellow color in fig.13.17 below.
(iii) So the quotient of the two functions will be:
$\frac{f(x)}{g(x)}~=~\frac{x^2 + 1}{x+4}$
• This is plotted in red color in fig.13.17 below.

Fig.13.17

2. Let us write the limit of f(x) as x approaches -2
• We get: $\lim_{x\rightarrow -2} g(x) = [(-2)^2 + 1] = (4+1) = 5$
• Indeed, we have a green filled circle at (-2,5)   
3. Let us write the limit of g(x) as x approaches -2
• We get: $\lim_{x\rightarrow -2} f(x) = (-2 + 4) = 2$
• Indeed, we have a yellow filled circle at (-2,2)
4. Let us write the limit of $\frac{f(x)}{g(x)}$ as x approaches -2
• We get: $\lim_{x\rightarrow -2}\left [\frac{f(x)}{g(x)}\right] = \frac{(-2)^2 +1}{-2+4} = \frac{5}{2} = 2.5$
• Indeed, we have a red filled circle at (-2,2.5)
5. Comparing the above steps (2), (3) and (4), we can write:
$$\lim_{x\rightarrow -2} \left [\frac{f(x)}{g(x)}\right] = 2.5 = \frac{\lim_{x\rightarrow -2} f(x)}{\lim_{x\rightarrow -2} g(x)} = \frac{5}{2}$$

---

• We have completed the examples for all the four cases.
    ♦ We have taken the limits at only a few x-values like -2 and 1.
    ♦ The reader may check other convenient x-values like -5, -0.5, 3.75 etc.,
• In the next section, we will see limits of polynomials and rational functions. 

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Chapter 13.3 - Algebra of Limits

In the previous section, we completed a discussion on the examples which demonstrated the basics of limits. In this section, we will see algebra of limits.

• In the previous sections, we have seen some examples in which:
   ♦ sum of the limits of two functions
   ♦ is equal to
   ♦ the limit of the sum of the two functions.
• For example:
$$\lim_{x\rightarrow 1} x^2~+~\lim_{x\rightarrow 1} x~=~\lim_{x\rightarrow 1} \left[x^2 + x \right]$$

---

• In the second chapter, we saw that:
   ♦ Two functions can be added together.
   ♦ One function can be subtracted from another function.
   ♦ A function can be multiplied by another function.
   ♦ A function can be divided by another function.
(Those details can be seen here)
◼ Now we will write a theorem of the limiting process related to addition, subtraction, multiplication and division of functions.
Theorem 1
This can be written in 4 steps:
Let f and g be two functions such that, both $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ exist.
Then:
1. Addition of individual limits is possible:
   ♦ Limit of the sum [f(x) + g(x)]
   ♦ is equal to
   ♦ the sum of the two individual limits.
• That is:
$$\lim_{x\rightarrow a} [f(x) + g(x)]=\lim_{x\rightarrow a} f(x)+\lim_{x\rightarrow a} g(x)$$
2. Difference of individual limits is possible:
   ♦ Limit of the difference [f(x) - g(x)]
   ♦ is equal to
   ♦ the difference of the two individual limits.
• That is:
$$\lim_{x\rightarrow a} [f(x) - g(x)]=\lim_{x\rightarrow a} f(x)-\lim_{x\rightarrow a} g(x)$$
3. Multiplication of individual limits is possible:
   ♦ Limit of the product [f(x) . g(x)]
   ♦ is equal to
   ♦ the product of the two individual limits.
• That is:
$$\lim_{x\rightarrow a} [f(x) \, . \, g(x)]=\lim_{x\rightarrow a} f(x) \, .\, \lim_{x\rightarrow a} g(x)$$
• Here there is special case also. It can be written in 2 steps:
(i) One of the function, say g(x) is a constant function $\lambda$.
   ♦ That is., $g(x) = \lambda$
   ♦ Where $\lambda$ is a real number.
(ii) Then we get:
$$\lim_{x\rightarrow a} [\lambda \, . \, f(x)]=\lambda \, .\, \lim_{x\rightarrow a} f(x)$$
4. Division of individual limits is possible:
   ♦ Limit of the quotient $\frac{f(x)}{g(x)}$
   ♦ is equal to
   ♦ the quotient of the two individual limits.
• That is:
$$\lim_{x\rightarrow a} \left[\frac{f(x)}{g(x)} \right] = \frac{\lim_{x\rightarrow a} f(x)}{\lim_{x\rightarrow a} g(x)}$$
• In this fraction, the denominator should not be equal to zero.

---

We will see the proof for the above theorem in higher classes. At present, we will see some examples.

Example for addition:
• This can be written in 5 steps:
1. Let us write the functions:
(i) Let the first function be: f(x) = x³ + 1
• This is plotted in green color in fig.13.13 below.
(ii) Let the second function be: g(x) = x² + 2x + 3
• This is plotted in yellow color in fig.13.13 below.
(iii) So the sum of the two functions will be:
f(x) + g(x) = x³ + x² + 2x + 4
• This is plotted in red color in fig.13.13 below.

Fig.12.13

2. Let us write the limit of f(x) as x approaches 1
• We get: $\lim_{x\rightarrow 1} f(x) = (1^3 + 1) = (1+1) = 2$
• Indeed, we have a green filled circle at (1,2)
3. Let us write the limit of g(x) as x approaches 1
• We get: $\lim_{x\rightarrow 1} g(x) = (1^2 + 2 × 1 + 3) = (1+2+3) = 6$
• Indeed, we have a yellow filled circle at (1,6)   
4. Let us write the limit of [f(x) + g(x)] as x approaches 1
• We get: $\lim_{x\rightarrow 1} [f(x) + g(x)] = (1^3 + 1^2 + 2 × 1 + 4) = (1+1+2+4) = 8$
• Indeed, we have a red filled circle at (1,8)
5. Comparing the above steps (2), (3) and (4), we can write:
$$\lim_{x\rightarrow 1} [f(x) + g(x)]= 8 = \lim_{x\rightarrow 1} f(x)+\lim_{x\rightarrow 1} g(x) = 2 + 6$$

Example for subtraction:
• This can be written in 5 steps:
1. Let us write the functions:
(i) Let the first function be: f(x) = x³ + 1
• This is plotted in green color in fig.13.14 below.
(ii) Let the second function be: g(x) = x² + 2x + 3
• This is plotted in yellow color in fig.13.14 below.
(iii) So the difference of the two functions will be:
f(x) - g(x) = x³ - x² - 2x - 2
• This is plotted in red color in fig.13.14 below.

Fig.13.14

2. Let us write the limit of f(x) as x approaches 1
• We get: $\lim_{x\rightarrow 1} f(x) = (1^3 + 1) = (1+1) = 2$
• Indeed, we have a green filled circle at (1,2)
3. Let us write the limit of g(x) as x approaches 1
• We get: $\lim_{x\rightarrow 1} g(x) = (1^2 + 2 × 1 + 3) = (1+2+3) = 6$
• Indeed, we have a yellow filled circle at (1,6)   
4. Let us write the limit of [f(x) - g(x)] as x approaches 1
• We get: $\lim_{x\rightarrow 1} [f(x) - g(x)] = (1^3 - 1^2 - 2 × 1 - 2) = (1-1-2-2) = -4$
• Indeed, we have a red filled circle at (1,-4)
5. Comparing the above steps (2), (3) and (4), we can write:
$$\lim_{x\rightarrow 1} [f(x) - g(x)]= -4 = \lim_{x\rightarrow 1} f(x)-\lim_{x\rightarrow 1} g(x) = 2 - 6$$

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In the next section, we will see examples for multiplication and division. 

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