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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21.1 - 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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