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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