In the previous section, we saw algebra of continuous functions. In this section, we will see some solved examples.
Solved Example 21.16
Prove that every rational function is continuous.
Solution:
1. Every rational function will be of the form: $f(x) = \frac{p(x)}{q(x)}$
♦ p(x) and q(x) are polynomial functions.
♦ Also q(x) should not be zero.
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3. In our present case, both numerator and denominator are polynomial functions. We have seen that all polynomial functions are continuous.
• So we can write: all rational functions are continuous.
4. We must consider the domain of a given rational function. That domain will not contain any real number which makes q(x) equal to zero.
• So for all values in the domain, the given rational function will be indeed continuous.
Solved Example 21.17
Discuss the continuity of the sine function.
Solution:
1. Fig.21.11 below shows the graph of the sine function f(x) = sin x.
 |
| Fig.21.11 |
2. An arbitrary point c is marked on the graph.
We want to find $\lim_{x\rightarrow c^{-}} f(x)$ and $\lim_{x\rightarrow c^{+}} f(x)$
3. First we will find $\lim_{x\rightarrow c^{-}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the left of c. We can write it as x = c−h
(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\lim_{x\rightarrow c^{-}} f(x)} & {~=~} &{\lim_{h\rightarrow 0} \sin (c-h)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\lim_{h\rightarrow 0} [\sin c \, \cos h - \cos c \, \sin h]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\lim_{h\rightarrow 0} [\sin c \, \cos h] ~-~\lim_{h\rightarrow 0} [\cos c \, \sin h]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\sin c \, \lim_{h\rightarrow 0} [\cos h] ~-~\cos c \, \lim_{h\rightarrow 0} [ \sin h]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\sin c \, \cos 0 ~-~\cos c \, \sin 0} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\sin c \, (1) ~-~\cos c \, (0)} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{\sin c} \\
\end{array}$
◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a−b) = sin a cos b − cos a sin b
• 5 (magenta color):
♦ From the graph of the cosine function, it is clear that, when the input value approaches zero from left or right, the limiting cosine value is 1.
♦ From the graph of the sine function, it is clear that, when the input value approaches zero from left or right, the limiting sine value is 0.
4. Next we will find $\lim_{x\rightarrow c^{+}} f(x)$. It can be done in 3 steps:
(i) Consider a point to the right of c. We can write it as x = c+h. This is shown in the graph below:
 |
| Fig.21.12 |
(ii) When x approaches c, h will approach zero.
(iii) So we can write:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}}
&{\lim_{x\rightarrow c^{+}} f(x)} & {~=~}
&{\lim_{h\rightarrow 0} \sin (c+h)} \\
{~\color{magenta}
2 } &{{}} &{{}} & {~=~}
&{\lim_{h\rightarrow 0} [\sin c \, \cos h + \cos c \, \sin h]}
\\
{~\color{magenta} 3 } &{{}} &{{}} &
{~=~} &{\lim_{h\rightarrow 0} [\sin c \, \cos h]
~+~\lim_{h\rightarrow 0} [\cos c \, \sin h]} \\
{~\color{magenta}
4 } &{{}} &{{}} & {~=~} &{\sin c \,
\lim_{h\rightarrow 0} [\cos h] ~+~\cos c \, \lim_{h\rightarrow 0} [ \sin
h]} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\sin c \, \cos 0 ~+~\cos c \, \sin 0} \\
{~\color{magenta} 6 } &{{}} &{{}} & {~=~} &{\sin c \, (1) ~+~\cos c \, (0)} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{\sin c} \\
\end{array}$
◼ Remarks:
• 2 (magenta color): Here we use the identity:
sin(a+b) = sin a cos b + cos a sin b
5. We see that, left side limit is equal to the right side limit. That means, limit exists.
6. We know that, the value of the function at x = c is:
f(c) = sin c
7. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = sin x is a continuous function.
In a similar way, we can prove that, the cosine function is also a continuous function. Steps are shown briefly below:
1. Left side limit can be calculated as:
◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a−b) = cos a cos b + sin a sin b
2. Right side limit can be calculated as:
◼ Remarks:
• 2 (magenta color): Here we use the identity:
cos(a+b) = cos a cos b − sin a sin b
3. We know that, the value of the function at x = c is:
f(c) = cos c
4. We can write: $\lim_{x\rightarrow c} f(x) ~=~f(c)$
So f(x) = cos x is a continuous function.
Solved Example 21.18
Discuss the continuity of the function f(x) = tan x.
Solution:
1. f(x) = tan x can be written as: $\rm{f(x) = \frac{\sin x}{\cos x}}$
2. From the algebra of limits, we know that:
If both numerator and denominator are continuous functions, then that fraction will be a continuous function.
3.
In our present case, we just saw in the above solved example that, both sine and cosine functions are continuous
functions.
• So we can write: f(x) = tan x is a continuous function.
4.
We must consider the domain of this function. That domain
will not contain any real number which makes 'cos x' equal to zero.
• So for all values in the domain, f(x) = tan x will be indeed continuous.
5. More details about the domain can be obtained from the graph below. It is the graph of f(x) = tan x.
 |
| Fig.21.13 |
• We see that:
When the input x value approaches $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the f(x) value approaches infinity. This is because, the denominator becomes smaller and smaller and approaches zero. When the input x values are exact $\rm{-{\frac{3 \pi}{2}},~-{\frac{\pi}{2}},~\frac{\pi}{2},~\frac{3 \pi}{2},\frac{5 \pi}{2}}$ etc., the denominator becomes zero. Division by zero will give a number which is not defined. So we must avoid these input values in the domain. In short, the domain should not contain those x values which are given by $\rm{(2n+1){\frac{\pi}{2}}}$, where n is any integer, +ve or -ve.
Now we will see continuity of composite functions. It can be written in 4 steps:
1. Consider the composite function (f∘g)(x). It can also be written as f(g(x)).
• We see that, the output of g is being used as the input for f.
2. Consider an arbitrary point c. When the input for g is c, the output will be g(c).
• Then the input for f will be g(c)
3. When we consider the two functions f and g together, we see that, two inputs are being made:
♦ c is the input for g
♦ g(c) is the input for f
4. Now we can write about the continuity of (f∘g).
♦ Suppose that, g is continuous at c.
♦ Also suppose that, f is continuous at g(c)
• Then we can write: (f∘g) is continuous at c.
We will see the proof in higher classes.
Solved Example 21.19
Show that the function defined by f(x) = sin (x2) is a continuous function.
Solution:
1. The given function is: f(x) = sin (x2)
2. We can write it as the composite of two functions: f(x) = g(h(x))
♦ Where h(x) = x2 and g(h(x)) = sin (h(x))
3. Consider h(x) = x2
• At any arbitrary point c, we know that, h will be continuous.
• The output of h at c is h(c) = c2.
4. So the input for g at c is c2.
• c2 is a real number. We know that sin x is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = sin (x2) is a continuous function.
Solved Example 21.20
Show that the function defined by f(x) = |1− x + |x|| is a continuous function.
Solution:
1. The given function is: f(x) = |1 − x + |x||
2. We can write it as the composite of two functions: f(x) = g(h(x))
♦ Where h(x) = 1 − x + |x| and g(h(x)) = |h(x)|
3. Consider h(x) = 1 − x + |x|
• This is the sum of two continuous functions:
(1−x) and |x|
♦ (1−x) is a polynomial function. So it is continuous.
♦ |x| is the modulus function. It is also continuous.
• Thus h, which is the sum of two continuous functions, will be continuous.
• In other words, at any arbitrary point c, h will be continuous.
• The output of h at c is h(c) = 1 − c + |c|.
4. So the input for g at c is (1 − c + |c|).
• (1 − c + |c|) is a real number. We know that g(x) = |x| is continuous for all real numbers.
• So g is continuous at h(c)
• Therefore, f(x) = g(h(x)) = |1 − x + |x|| is a continuous function.
The link to a few more solved examples is given below:
In the next section, we will see differentiability.
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