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.

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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.

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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.

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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.

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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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21.2 - Solved Examples on Continuity

In the previous section, we saw continuous functions. We also saw some solved examples and the concept of infinity. In this section, we will see a few more solved examples.

Solved example 21.10
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x+2,  & \text{if}~x\le 1 \\[1.5ex] x-2, & \text{if}~x>1 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than or equal to 1, we must use the segment f(x) = x+2
(ii) When input x is greater than 1, we must use the segment f(x) = x−2
• So x = 1 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 1
   ♦ Case 2, where input x is greater than 1
   ♦ Case 3, where input x is equal to 1
These three cases will cover all real numbers.

Case 1: x <1
1. Consider any arbitrary point c such that c < 1. Let us check whether the function is continuous at c.
Since c < 1, we must use the segment f(x) = x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c+2
• 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 less than 1. So it can be any real number less than 1.
• Therefore we can write:
The given function is continuous at every real number less than 1.      

Case 2: x > 1
1. Consider any arbitrary point c such that c > 1. Let us check whether the function is continuous at c.
Since c > 1, we must use the segment f(x) = x−2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c−2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c−2
• 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 greater than 1. So it can be any real number greater than 1.
• Therefore we can write:
The given function is continuous at every real number greater than 1.

Case 3: x = 1
1. Here we cannot consider an arbitrary point. We must check the continuity at the exact point where x = 1.
2. Applying condition (i):
Since x = 1 is a critical point, we must check both left side limit and right side limit.
◼ Checking left side limit:
• While checking left side limit, all input x values are less than 1. So we must use the segment f(x) = x+2
• Therefore, $\lim_{x\rightarrow 1^{−}} f(x) = (1+2) = 3$
◼ Checking right side limit:
• While checking right side limit, all input x values are greater than 1. So we must use the segment f(x) = x-2
• Therefore, $\lim_{x\rightarrow 1^{+}} f(x) = (1-2) = -1$
◼ The two limits are not the same. So we can write:
$\lim_{x\rightarrow 1} f(x)$ does not exist.
• So this function does not satisfy the first condition at x =1.
3. We can establish continuity only if both conditions are satisfied. Therefore, this function is not continuous at x = 1.

Conclusion:
From cases 1 and 2, we established continuity at all points other than 1.
• So we can write:
For the given function, x = 1 is the only point of discontinuity.

The above steps will become more clear if we use a graph. It is shown in fig.21.6 below:

Fig.21.6

1. For case 1, we use the left segment. We see that, for any point where x<1, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>1, the function is continuous.
3. For case 3, we use the two points (1,3) and (1,-1).
   ♦ Left side limit is indicated by (1,3).
   ♦ Right side limit is indicated by (1,-1).
4. We see that, the graph cannot be drawn in a single stroke. At (1,3), we have to lift the pen from the plane of the paper. Then continue drawing from (1,-1).
5. Note the type of circles used for marking the two points.
   ♦ (1,3) is marked with a filled circle.
   ♦ (1,-1) is marked with an ordinary circle.

Solved example 21.11
Find all the points of discontinuity of the function f defined by
$f(x) = \begin{cases} x+2,  & \text{if}~x< 1
 \\[1.5ex] 0, & \text{if}~x=1
\\[1.5ex] x-2, & \text{if}~x>1
\end{cases}$
Solution:
• The given function has three segments.
(i) When input x is less than 1, we must use the segment f(x) = x+2
(ii) When input x is equal to 1, we must use the segment f(x) = 0
(iii) When input x is greater than 1, we must use the segment f(x) = x−2
• So x = 1 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 1
   ♦ Case 2, where input x is greater than 1
   ♦ Case 3, where input x is equal to 1
These three cases will cover all real numbers.

Case 1: x <1
1. Consider any arbitrary point c such that c < 1. Let us check whether the function is continuous at c.
Since c < 1, we must use the segment f(x) = x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c+2
• 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 less than 1. So it can be any real number less than 1.
• Therefore we can write:
The given function is continuous at every real number less than 1.      

Case 2: x > 1
1. Consider any arbitrary point c such that c > 1. Let us check whether the function is continuous at c.
Since c > 1, we must use the segment f(x) = x−2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c−2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c−2
• 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 greater than 1. So it can be any real number greater than 1.
• Therefore we can write:
The given function is continuous at every real number greater than 1.

Case 3: x = 1
1. Here we cannot consider an arbitrary point. We must check the continuity at the exact point where x = 1.
2. Applying condition (i):
Since x = 1 is a critical point, we must check both left side limit and right side limit.
◼ Checking left side limit:
• While checking left side limit, all input x values are less than 1. So we must use the segment f(x) = x+2
• Therefore, $\lim_{x\rightarrow 1^{−}} f(x) = (1+2) = 3$
◼ Checking right side limit:
• While checking right side limit, all input x values are greater than 1. So we must use the segment f(x) = x-2
• Therefore, $\lim_{x\rightarrow 1^{+}} f(x) = (1-2) = -1$
◼ The two limits are not the same. So we can write:
$\lim_{x\rightarrow 1} f(x)$ does not exist.
• So this function does not satisfy the first condition at x =1.
3. We can establish continuity only if both conditions are satisfied. Therefore, this function is not continuous at x = 1.

Conclusion:
From cases 1 and 2, we established continuity at all points other than 1.
• So we can write:
For the given function, x = 1 is the only point of discontinuity.

• It may be noted that, in the previous example,
$\lim_{x\rightarrow 1^{−}} f(x) ~\ne~ \lim_{x\rightarrow 1^{+}} f(x)$
But $\lim_{x\rightarrow 1^{−}} f(x) ~=~ f(1)$
• In the present example:
$\lim_{x\rightarrow 1^{−}} f(x) ~\ne~ \lim_{x\rightarrow 1^{+}} f(x) ~\ne~ f(1)$.
This is because, for the present example, it is given that: f(1) = 0.

The above steps will become more clear if we use a graph. It is shown in fig.21.7 below:

Fig.21.7

1. For case 1, we use the left segment. We see that, for any point where x<1, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>1, the function is continuous.
3. For case 3, we use the two points (1,3) and (1,-1).
   ♦ Left side limit is indicated by (1,3).
   ♦ Right side limit is indicated by (1,-1).
4. We see that, the graph cannot be drawn in a single stroke. At (1,3), we have to lift the pen from the plane of the paper. Then continue drawing from (1,-1).
5. Note the type of circles used for marking the three points.
   ♦ (1,3) is marked with a ordinary circle.
   ♦ (1,-1) is marked with an ordinary circle.
   ♦ (1,0) is marked with a filled circle.

Solved example 21.12
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x+2,  & \text{if}~x < 0 \\[1.5ex] -x+2, & \text{if}~x>0 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than 0, we must use the segment f(x) = x+2
(ii) When input x is greater than 0, we must use the segment f(x) = −x+2
• So x = 0 is a critical point. We must use this point for analysis. We must consider two cases:
   ♦ Case 1, where input x is less than zero.
   ♦ Case 2, where input x is greater than zero.
These two cases will cover all real numbers except zero.

Case 1: x < 0
1. Consider any arbitrary point c such that c < 0. Let us check whether the function is continuous at c.
Since c < 0, we must use the segment f(x) = x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c+2
• 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 less than 0. So it can be any real number less than 0.
• Therefore we can write:
The given function is continuous at every real number less than 0.      

Case 2: x > 0
1. Consider any arbitrary point c such that c > 0. Let us check whether the function is continuous at c.
Since c > 0, we must use the segment f(x) = −x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: −c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = −c+2
• 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 greater than 0. So it can be any real number greater than 0.
• Therefore we can write:
The given function is continuous at every real number greater than 0.

Conclusion:
From cases 1 and 2, we established continuity at all points other than zero.
• We need not consider the continuity at zero because, it is not defined in the given function.
• Domain of the given function is: R − {0}
• So we can write:
Since f is continuous at all points in the domain, it is a continuous function.

The above steps will become more clear if we use a graph. It is shown in fig.21.8 below:

Fig.21.8

1. For case 1, we use the left segment. We see that, for any point where x<0, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>0, the function is continuous.
3. We need not consider the case when x = 0.
4. Note the type of circles used for marking the point where x = 0. It is an ordinary circle. The ordinary circle indicates that, the point is not included in the graph.
5. We see that, the graph cannot be drawn in a single stroke. At x=0, we have to lift the pen from the plane of the paper. But x=0 is not defined. Zero is not present in the domain.
• So we can write:
The given function f is continuous for all points in the domain of f.

Solved example 21.13
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x,  & \text{if}~x \ge 0 \\[1.5ex] x^2, & \text{if}~x < 0 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than zero, we must use the segment f(x) = x².
(ii) When input x is greater than or equal to zero, we must use the segment f(x) = x.
• So x = 0 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 0
   ♦ Case 2, where input x is greater than 0
   ♦ Case 3, where input x is equal to 0
These three cases will cover all real numbers.

Case 1: x < 0
1. Consider any arbitrary point c such that c < 0. Let us check whether the function is continuous at c.
Since c < 0, we must use the segment f(x) = x².
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c².
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c².
• 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 less than 0. So it can be any real number less than 0.
• Therefore we can write:
The given function is continuous at every real number less than zero.      

Case 2: x > 0
1. Consider any arbitrary point c such that c > 0. Let us check whether the function is continuous at c.
Since c > 0, we must use the segment f(x) = x
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 greater than 0. So it can be any real number greater than 0.
• Therefore we can write:
The given function is continuous at every real number greater than zero.

Case 3: x = 0
1. Here we cannot consider an arbitrary point. We must check the continuity at the exact point where x = 0.
2. Applying condition (i):
Since x = 0 is a critical point, we must check both left side limit and right side limit.
◼ Checking left side limit:
• While checking left side limit, all input x values are less than 0. So we must use the segment f(x) = x².
• Therefore, $\lim_{x\rightarrow 1^{−}} f(x) = (0^2) = 0$
◼ Checking right side limit:
• While checking right side limit, all input x values are greater than 0. So we must use the segment f(x) = x
• Therefore, $\lim_{x\rightarrow 1^{+}} f(x) = 0$
◼ The two limits are the same. So we can write:
$\lim_{x\rightarrow 0} f(x)$ exists.
• So this function satisfies the first condition at x = 0.
3. Applying condition (ii):
When x = 0, we must use the segment f(x) = x.
So we get: f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = 0.

Conclusion:
From cases 1, 2 and 3, we established continuity at all real numbers.
• The domain of the function is R. So we can write:
The given function f, is continuous at all points in the domain of f. So it is a continuous function.

The above steps will become more clear if we use a graph. It is shown in fig.21.9 below:

Fig.21.9

1. For case 1, we use the left segment. We see that, for any point where x<0, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>0, the function is continuous.
3. For case 3, we use the point (0,0).
   ♦ Left side limit is zero.
   ♦ Right side limit is zero.
4. We see that, the graph can be drawn in a single stroke. We do not have to lift the pen from the plane of the paper. 

Solved example 21.13
Show that every polynomial function is continuous.
Solution:
1. A function p is polynomial function if it is defined by:
p(x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ.
   ♦ a, a₁, a₂ etc., must be real numbers.
   ♦ n must be a natural number.
(Details can be seen here)
2. Many functions that we analyzed in this chapter are polynomial functions.
• f(x) = x is a polynomial function.
   ♦ Here a₀ = 0 and a₁ = 1
• f(x) = x² is a polynomial function.
   ♦ Here a₀ = 0, a₁ = 0 and a₂ = 1
3. Some graphs of polynomial functions are shown below:

Fig.21.10

4. We can draw the graphs of polynomial functions without lifting the pen from the plane of the paper.
5. We can consider all polynomial functions as continuous functions.
• We will see detailed proof in the next section.

Solved example 21.14
Find all the points of discontinuity of the greatest integer function defined by f(x)=$\mathbf\small{\rm{\left\lfloor x\right\rfloor }}$, where $\mathbf\small{\rm{\left\lfloor x\right\rfloor }}$ denotes the greatest integer less than or equal to x.
Solution:
• Details about the greatest integer function can be seen in section 2.6.
• We must consider two cases:
   ♦ Case 1, where input x is not an integer.
   ♦ Case 2, where input x is an integer
These two cases will cover all real numbers.

Case 1: x not an integer.
1. Consider any arbitrary point c such that c is not an integer. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is:
$\mathbf\small{\rm{\left\lfloor c \right\rfloor }}$ = The greatest integer less than c.
• For example,
    ♦ if c = 3.58, then the limiting value = $\mathbf\small{\rm{\left\lfloor 3.58 \right\rfloor }}$ = 3 
    ♦ if c = −1.5, then the limiting value = $\mathbf\small{\rm{\left\lfloor -1.5 \right\rfloor }}$ = −2 
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = $\mathbf\small{\rm{\left\lfloor c \right\rfloor }}$ = The greatest integer less than 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 which is not an integer. So it can be any real number which is not an integer.
• Therefore we can write:
The given function is continuous at every real number which is not an integer.      

Case 2: x is an integer
1. Consider any arbitrary point c such that c is an integer. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is to be calculated.
We need to find the left side limit and right side limit.
    ♦ The left side limit will be (c-1).
    ♦ The right side limit will be c.
• For example,
    ♦ if c = 3, then the L.S.L = 2 and R.S.L = 3  
    ♦ if c = −2, then the L.S.L = −3 and R.S.L = −2  
(see graph in the fig.2.17 in section 2.6)
• It is clear that, $\lim_{x\rightarrow c} f(x)$ does not exist. So first condition is not satisfied.
3. We can establish continuity at a point only if both conditions are satisfied at that point.
4. c was taken as an arbitrary point which is an integer. So it can be any real number which is an integer.
• Therefore we can write:
The given function is not continuous at integers.

Conclusion:
• From cases 1 and 2, we see that,
    ♦ f(x) = is continuous at all real numbers which are not integers.
    ♦ But it is not continuous at every real number, which is an integer.

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In the next section, we will see algebra of continuous functions.

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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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Chapter 21 - Continuity and Differentiability

In the previous section, we completed a discussion on determinants. In this section, we will see Continuity and Differentiability.

• We have already learned about limits and derivatives in a previous chapter. The reader is advised to revise those topics thoroughly, before attempting our present chapter.

• Consider fig.13.5 that we saw in that previous chapter. We can use that figure to learn the basics about continuity. It can be written in 4 steps:
1. In fig.13.5, we wanted the limit at x = 0.
• We saw that:
   ♦ The left side limit is 1.
   ♦ The right side limit is 2.
2. So, the limit at x = 0 does not exist.
3. In general, if at a point c, if the left side limit is not equal to the right side limit, we say that, the function is discontinuous at c. Also, c is a point of discontinuity.
4. Note that, in fig.13.5, at x = 0, we cannot draw the graph without lifting the pen from the plane of the paper. So there indeed, is a "discontinuity".

Let us see another example. It can be written in 5 steps:
1. Consider fig.13.7 that we saw in that previous chapter. We wanted the limit at x = 1.
• We saw that:
   ♦ The left side limit is 3.
   ♦ The right side limit is also 3.
2. So, the limit at x = 1 exists.
• But value of the function at x= 1, is not 3.
3. So in some cases,
   ♦ The left side and right side limits may be same at a point c.
   ♦ But value of the function at c, may be different from that limiting value.
4. The situation mentioned in the above step (3), is also a discontinuity. We say that, the function is discontinuous at c. And, c is a point of discontinuity.
5. Note that, in fig.13.7, at x = 1, we cannot draw the graph without lifting the pen from the plane of the paper. So there indeed, is a "discontinuity".

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• We have seen two examples which were related to two different situations. Those two  different situations gave us a basic idea about discontinuity. So now we can write "opposite situations" which will give us a basic idea about continuity. It can be written in 4 steps:
1. A function is said to be continuous at a point c, if two conditions are satisfied:
Condition (i):
   ♦ The left side limit at c
   ♦ is equal to
   ♦ The right side limit at c.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c^{-}} f(x) ~=~ \lim_{x\rightarrow c^{+}} f(x)$

Condition (ii):
   ♦ Limiting value at c
   ♦ is equal to
   ♦ The value of the function at c.

2. It is possible to write the conditions in a simpler way.
• Consider the first condition that we wrote above. If this condition is satisfied, it means that, limit at c exists. So we can write:
◼  A function is said to be continuous at a point c, if two conditions are satisfied:
Condition (i):
Limit at c exists.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c} f(x)$ exists.

Condition (ii):
   ♦ Limiting value at c
   ♦ is equal to
   ♦ The value of the function at c.
• Using symbols, we can write this condition as:
$\lim_{x\rightarrow c} f(x)~=~f(c)$.

3. We have seen numerous solved examples where we checked "whether limit at any given point c exists".
• All we need to do is: Find the limit at c.
• If the limiting value is in the form 0/0 or "a division by zero", we concluded that, the limit does not exist.
• Also note that, "finding the limit at c" is essential because, only then we will be able to apply the second condition.

4. When the two conditions are satisfied, we will be able to draw the graph in a single stroke, with out lifting the pen from the plane of the paper.

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Let us see some solved examples:

Solved example 21.1
Check the continuity of the function given by f(x) = 2x + 3 at x = 1.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 1 is:
2(1) + 3 = 5
• It is clear that, $\lim_{x\rightarrow 1} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(1) = 2(1) + 3 = 5
• We see that: $\lim_{x\rightarrow 1} f(x)~=~f(1)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 1.

Solved example 21.2
Check the continuity of the function given by f(x) = x² at x = 0.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 0 is:
(0)² = 0
• It is clear that, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(0) = (0)² = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(0)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 0.

Solved example 21.3
Check the continuity of the function given by f(x) = |x| at x = 0.
Solution:
1. Applying condition (i):
This condition can be applied with greater clarity, if we draw the graph. It is shown in fig.21.1 below:

Fig.21.1

• When x approaches zero from the left, f(x) approaches zero. That is: $\lim_{x\rightarrow 0^{-}} f(x) = 0$
• When x approaches zero from the right, f(x) approaches zero. That is: $\lim_{x\rightarrow 0^{+}} f(x) = 0$
• Both left side and right side limits are the same. That means, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
• From the graph, we have: f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(0)$
• So second condition is also satisfied.
3. Since both conditions are satisfied, the given function is continuous at x = 0.

Solved example 21.4
Show that the function f given by
$f(x) = \begin{cases} x^3+3,  & \text{if}~x\ne0 \\[1.5ex] 1, & \text{if}~x=0 \end{cases}$
is not continuous at x = 0.
Solution:
1. Applying condition (i):
• Limiting value of f(x) at x = 0 is:
(0)³ + 3 = 3
• It is clear that, $\lim_{x\rightarrow 0} f(x)$ exists. So first condition is satisfied.
2. Applying condition (ii):
f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~\ne~f(0)$
• So second condition is not satisfied.
3. Since both conditions are not satisfied, the given function is not continuous at x = 0.

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In the next section, we will see continuous functions.

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20.16 - Miscellaneous Examples

In the previous section, we completed a discussion on determinants. We saw a solved example also. In this section, we will see some miscellaneous examples.

Solved example 20.27
If a, b, c are positive and unequal, show that the value of the determinant $\Delta ~=~\left |\begin{array}{r}                           
a    &{    b    }    &{    c    }    \\
b    &{    c    }    &{    a    }    \\
c    &{    a    }    &{    b    }    \\
\end{array}\right |
$ is negative.
Solution:
1. First we will simplify the given determinant:

◼ Remarks:
• 2 (magenta color): Apply R₁ → R₁ + R₂.
• 3 (magenta color): Apply R₁ → R₁ + R₃.
• 4 (magenta color): Apply C₃ → C₃ − C₁.
• 5 (magenta color): Apply C₂ → C₂ − C₁.
• 6 (magenta color): Expand along R₁.

2. Consider the above result. There are three terms:
(i) −(1/2)
(ii) (a+b+c)
(iii) (a−b)² + (a−c)² + (b−c)².

• Given that: a, b, c are +ve and unequal.
• So (ii) and (iii) cannot become -ve.
• Therefore, due to the presence of −(1/2), the result as a whole will become -ve.

Solved example 20.28
If a, b, c are in A.P, find the value of
$\Delta ~=~\left |\begin{array}{r}                           
2y+4    &{    5y+7    }    &{    8y+a    }    \\
3y+5    &{    6y+8    }    &{    9y+b    }    \\
4y+6    &{    7y+9    }    &{    10y+c    }    \\
\end{array}\right |
$.
Solution:

◼ Remarks:
• 2 (magenta color): Apply R₃ → R₃ − R₂.
• 3 (magenta color): Apply R₂ → R₂ − R₁.
• 4 (magenta color): Apply R₃ → R₃ − R₂.
• 5 (magenta color): Since, a, b, c are in A.P, we can put 2b = a+c.
• 6 (magenta color): All elements of R₃ are zeroes. So the value of the determinant is zero.

Solved example 20.29
Show that
$\Delta ~=~\left |\begin{array}{r}                         
(y+z)^2    &{    xy    }    &{    zx    }    \\
xy    &{    (x+z)^2    }    &{    yz    }    \\
xz    &{    yz    }    &{    (x+y)^2    }    \\
\end{array}\right | ~=~2xyz (x+y+z)^3
$.
Solution:

◼ Remarks:
• 2 (magenta color):
    ♦ Multiply R₁ by x
    ♦ Multiply R₂ by y
    ♦ Multiply R₃ by z
To balance these multiplications, the whole determinant should be multiplied by (1/xyz)
• 3 (magenta color):
Take out the common factors:
    ♦ x from C₁
    ♦ y from C₂
    ♦ z from C₃
• 4 (magenta color):
Apply two operations:
    ♦ C₂ → C₂ − C₁
    ♦ C₃ → C₃ − C₁
• 5 (magenta color):
    ♦ Apply the identity: a² − b² = (a+b)(a-b).
    ♦ This is applied to C₂ and C₃.
• 6 (magenta color): Take out (x+y+z) from C₁ and C₂.
• 7 (magenta color):
Apply R₁ → R₁ − R₂
• 8 (magenta color):
Apply R₁ → R₁ − R₃
• 9 (magenta color):
Apply C₂ → C₂ + (1/y)C₁
• 10 (magenta color):
Apply C₃ → C₃ + (1/z)C₁
• 11 (magenta color):
Expand along R₁.

Solved example 20.30
Use product
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$
to solve the system of equations
x - y + 2z = 1
2y − 3z = 1
3x − 2y + 4z = 2
Solution:
1. Use matrix multiplication to find the product.
• We get:
$\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ] \left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~ = \left [\begin{array}{r}                         
1    &{    0    }    &{   0    }    \\
0    &{    1    }    &{   0    }    \\
0    &{    0    }    &{   1    }    \\
\end{array}\right ]
$

2. The product is an identity matrix. So it is clear that:
Inverse of $\left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ]$ is $\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]$

3. The given system can be written in the form AX = B.
$A = \left [\begin{array}{r}                         
1    &{    -1    }    &{    2    }    \\
0    &{    2    }    &{    -3    }    \\
3    &{    -2    }    &{   4    }    \\
\end{array}\right ],~X = \left[\begin{array}{r}       
x        \\
y        \\
z        \\
\end{array}\right]~~ \text{and}~~B = \left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]
$

4. So X = A⁻¹ B.
• Check whether A⁻¹ exists:
We have already obtained the inverse. Therefore, A⁻¹ exists.

5. Use matrix multiplication to find A⁻¹B.
• We get: X = A⁻¹ B =
$\left [\begin{array}{r}                         
-2    &{    0    }    &{    1    }    \\
9    &{    2    }    &{    -3    }    \\
6    &{    1    }    &{   -2    }    \\
\end{array}\right ]~\left[\begin{array}{r}                           
1        \\
1        \\
2        \\
\end{array}\right]~ = \left[\begin{array}{r}                        0        \\
5        \\
3        \\
\end{array}\right]
$

6. So the solution is: x = 0, y = 5 and z = 3

Solved example 20.31
Prove that
$ \Delta ~=~ \left |\begin{array}{r}                         
a+bx    &{    c+dx    }    &{    p+qx    }    \\
ax+b    &{    cx+d    }    &{    px+q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right | ~=~ (1 - x^2) \left |\begin{array}{r}                         
a    &{    c    }    &{    p    }    \\
b    &{    d    }    &{    q    }    \\
u    &{    v    }    &{   w    }    \\
\end{array}\right |$
Solution:
1. First we will split the given matrix by applying property V.

◼ Remarks:
• 2 (magenta color):
We split R₁ so that, a, c and p are obtained in the first row. These are the elements that we want in the R₁ of the final result.

2. Now we simplify |A|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |A|.
• 3 (magenta color): Consider the first determinant in (2). Every element in R₂ is proportional to the corresponding elements in R₁, by the same ratio 'x'. So this determinant becomes zero.

3. Next we simplify |B|:

◼ Remarks:
• 2 (magenta color): We split R₂ of |B|.
• 3 (magenta color): Consider the second determinant in (2). Every element in R₁ is proportional to the corresponding elements in R₂, by the same ratio 'x'. So this determinant becomes zero.
• 4 (magenta color): We take out the common factor 'x' from R₁ and R₂.
• 5 (magenta color): We want the elements a, c and p in R₁. So we interchange R₁ and R₂. The sign of the determinant will change when the two rows are interchanged.

4. Finally we add |A| and |B|. We get:

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The link below gives a few more examples:

Miscellaneous Examples

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In the next section, we will see Continuity and Differentiability.

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