Find the points of discontinuity, if any, of the following functions:

Solution:

(i)

When $x \neq 1$, then

$f(x)=x^{3}-x^{2}+2 x-2$

We know that a polynomial function is everywhere continuous.

So, $f(x)=x^{3}-x^{2}+2 x-2$ is continuous at each $x \neq 1$.

At $x=1$, we have

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left((1-h)^{3}-(1-h)^{2}+2(1-h)-2\right)=1-1+2-2=0$

$(\mathrm{RHL}$ at $x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left((1+h)^{3}-(1+h)^{2}+2(1+h)-2\right)=1-1+2-2=0$

Also, $f(1)=4$

$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \neq f(1)$

Thus, $f(x)$ is discontinuous at $x=1$.

Hence, the only point of discontinuity for $f(x)$ is $x=1$.

(ii)

Given: $f(x)=\left\{\begin{array}{l}\frac{x^{4}-16}{x-2}, \text { if } x \neq 2 \\ 16, \text { if } x=2\end{array}\right.$

When $x \neq 2$, then

$f(x)=\frac{x^{4}-16}{x-2}=\frac{x^{4}-2^{4}}{x-2}=\frac{\left(x^{2}+4\right)(x-2)(x+2)}{x-2}=\left(x^{2}+4\right)(x+2)$

We know that a polynomial function is everywhere continuous.

Therefore, the functions $\left(x^{2}+4\right)$ and $(x+2)$ are everywhere continuous.

So, the product function $\left(x^{2}+4\right)(x+2)$ is everywhere continuous.

Thus, $f(x)$ is continuous at every $x \neq 2$.

At $x=2$, we have

$(\mathrm{LHL}$ at $x=2)=\lim _{x \rightarrow 2} f(x)=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}\left[(2-h)^{2}+4\right](2-h+2)=8(4)=32$

(RHL at $x=2$ ) $=\lim _{x \rightarrow 2^{+}} f(x)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0}\left[(2+h)^{2}+4\right](2+h+2)=8(4)=32$

Also, $f(2)=16$

$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x) \neq f(2)$

Thus, $f(x)$ is discontinuous at $x=2$.

Hence, the only point of discontinuity for $f(x)$ is $x=2$.

(iii)

When $x<0$, then

$f(x)=\frac{\sin x}{x}$

We know that $\sin x$ as well as the identity function $x$ are everywhere continuous.

So, the quotient function $\frac{\sin x}{x}$ is continuous at each $x<0$.

When $x>0$, then

$f(x)=2 x+3$, which is a polynomial function. Therefore, $f(x)$ is continuous at each $x>0$.

Now,

Let us consider the point $x=0$.

Given: $f(x)=\left\{\begin{array}{l}\frac{\sin x}{x}, \text { if } x<0 \\ 2 x+3, \text { if } x \geq 0\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{\sin (-h)}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{\sin (h)}{h}\right)=1$

(RHL at $x=0$ ) $=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(2 h+3)=3$

$\therefore \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$

Thus, $f(x)$ is discontinuous at $x=0$.

Hence, the only point of discontinuity for $f(x)$ is $x=0$.

(iv)

When $x \neq 0$, then

$f(x)=\frac{\sin 3 x}{x}$

We know that $\sin 3 x$ as well as the identity function $x$ are everywhere continuous.

So, the quotient function $\frac{\sin 3 x}{x}$ is continuous at each $x \neq 0$.

Let us consider the point $x=0$.

Given: $f(x)=\left\{\begin{array}{l}\frac{\sin 3 x}{x}, \text { if } x \neq 0 \\ 4, \text { if } x=0\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{\sin (-3 h)}{-h}\right)=\lim _{h \rightarrow 0}\left(\frac{3 \sin (3 h)}{3 h}\right)=3$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}\left(\frac{\sin (3 h)}{h}\right)=\lim _{h \rightarrow 0}\left(\frac{3 \sin (3 h)}{3 h}\right)=3$

Also, $f(0)=4$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

Thus, $f(x)$ is discontinuous at $x=0$.

Hence, the only point of discontinuity for $f(x)$ is $x=0$.

(v)

When $x \neq 0$, then

$f(x)=\frac{\sin x}{x}+\cos x$

We know that $\sin x$ as well as the identity function $x$ both are everywhere continuous.

So, the quotient function $\frac{\sin x}{x}$ is continuous at each $x \neq 0$.

Also, $\cos x$ is everywhere continuous.

Therefore, $\frac{\sin x}{x}+\cos x$ is continuous at each $x \neq 0$.

Let us consider the point $x=0$.

Given: $f(x)=\left\{\begin{array}{l}\frac{\sin x}{x}+\cos x, \text { if } x \neq 0 \\ 5, \text { if } x=0\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{\sin (-h)}{-h}+\cos (-h)\right)=\lim _{h \rightarrow 0}\left(\frac{\sin (-h)}{-h}\right)+\lim _{h \rightarrow 0} \cos (-h)=1+1=2$

$($ RHL at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}\left(\frac{\sin (h)}{h}+\cos (h)\right)=\lim _{h \rightarrow 0}\left(\frac{\sin (h)}{h}\right)+\lim _{h \rightarrow 0} \cos (h)=1+1=2$

Also, $f(0)=5$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

Thus, $f(x)$ is discontinuous at $x=0$.

Hence, the only point of discontinuity for $f(x)$ is $x=0$.

(vi)

When $x \neq 0$, then

$f(x)=\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x}$

We know that $x^{4}+x^{3}+2 x^{2}$ is a polynomial function which is everywhere continuous.

Also, $\tan ^{-1} x$ is everywhere continuous.

So, the quotient function $\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x}$ is continuous at each $x \neq 0$.

Let us consider the point $x=0$.

Given: $f(x)=\left\{\begin{array}{l}\frac{x^{4}+x^{3}+2 x^{2}}{\tan ^{-1} x}, \text { if } x \neq 0 \\ 10, \text { if } x=0\end{array}\right.$

We have

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left(\frac{(-h)^{4}+(-h)^{3}+2(-h)^{2}}{\tan ^{-1}(-h)}\right)=\lim _{h \rightarrow 0}\left(\frac{(h)^{3}-(h)^{2}+2(h)}{-\frac{\tan ^{-1}(h)}{h}}\right)=\frac{0}{(-1)}=0$

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}\left(\frac{(h)^{4}+(h)^{3}+2(h)^{2}}{\tan ^{-1}(h)}\right)=\lim _{h \rightarrow 0}\left(\frac{(h)^{3}+(h)^{2}+2(h)}{\frac{\tan ^{-1}(h)}{h}}\right)=\frac{0}{1}=0$

Also, $f(0)=10$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x) \neq f(0)$

Thus, $f(x)$ is discontinuous at $x=0$.

Hence, the only point of discontinuity for $f(x)$ is $x=0$.

(vii)

Given: $f(x)=\left\{\begin{array}{l}\frac{e^{x}-1}{\log _{e}(1+2 x)}, \text { if } x \neq 0 \\ 7, \text { if } x=0\end{array}\right.$

We have

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{e^{x}-1}{\log _{e}(1+2 x)}=\lim _{x \rightarrow 0} \frac{\left(\frac{c^{x}-1}{x}\right)}{\left(\frac{2 \log _{e}(1+2 x)}{2 x}\right)}=\frac{1}{2} \times \frac{\lim _{x \rightarrow 0}\left(\frac{c^{x}-1}{x}\right)}{\lim _{x \rightarrow 0}\left(\frac{\log _{e}(1+2 x)}{2 x}\right)}=\frac{1}{2}$

It is given that $f(0)=7$

$\Rightarrow \lim _{x \rightarrow 0} f(x) \neq f(0)$

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

(viii)

When $x>1$, then

$f(x)=|x-3|$

Since modulus function is a continuous function, $f(x)$ is continuous for each $\underline{\underline{X}}>1$.

When $x<1$, then

$f(x)=\frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}$

Since, $x^{2} \& 3 x$ are continuous being polynomial functions, $\frac{x^{2}}{4} \& \frac{3 x}{2}$ will also be continuous. Also, $\frac{13}{4}$ is continuous being a polynomial function.

$\Rightarrow \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}$ is continuous for each $x<1$.

$\Rightarrow f(x)$ is continuous for each $x<1$

At $x=1$, we have

$(\mathrm{LHL}$ at $x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left[\frac{(1-h)^{2}}{4}-\frac{3(1-h)}{2}+\frac{13}{4}\right]=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$

(RHL at $x=1$ ) $=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[|1+h-3|]=|-2|=2$

Also, $f(1)=|1-3|=|-2|=2$

Thus, $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$

Hence, $f(x)$ is continuous at $x=1$.

Thus, the given function is nowhere discontinuous.

(ix)

At $x \leq-3$, we have

$f(x)=|x|+3$

Since modulus function and constant function are continuous, $f(x)=|x|+3$ is continuous for each $x \leq-3$.

At $-3

$f(x)=-2 x$

Since polynomial function is continuous and constant function is continuous, $f(x)=-2 x$ is continuous for each $-3

At $x>3$, we have

 

$f(x)=6 x+2$

Since polynomial function is continuous and constant function is continuous, $f(x)=6 x+2$ is continuous for each $x>3$.

Now, we check the continuity of the function at the point $x=3$.

We have

$(\mathrm{LHL}$ at $x=3)=\lim _{x \rightarrow 3} f(x)=\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}-2(3-h)=-6$

(RHL at $x=3$ ) $=\lim _{x \rightarrow 3^{+}} f(x)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} 6(3+h)+2=20$

 

$\Rightarrow \lim _{x \rightarrow 3^{-}} f(x) \neq \lim _{x \rightarrow 3^{+}} f(x)$

Hence, the only point of discontinuity of the given function is $x=3$

$(x)$

Given: $f(x)= \begin{cases}x^{10}-1, & \text { if } x \leq 1 \\ x^{2}, & \text { if } x>1\end{cases}$

The given function $f$ is defined at all the points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<1$, then $f(c)=c^{10}-1$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{10}-1\right)=c^{10}-1$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<1$

Case II:

If $c=1$, then the left hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(x^{10}-1\right)=1^{10}-1=1-1=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}\left(x^{2}\right)=1^{2}=1$

It is observed that the left and right hand limit of $f$ at $x=1$ do not coincide.

Therefore, $f$ is not continuous at $x=1$

Case III:

If $c>1$, then $f(c)=c^{2}$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2}\right)=c^{2}$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observation, it can be concluded that $x=1$ is the only point of discontinuity of $f$.

(xi) The given function is $f(x)= \begin{cases}2 x, & \text { if } x<0 \\ 0, & \text { if } 0 \leq x \leq 1 \\ 4 x, & \text { if } x>1\end{cases}$

The given function is defined at all points of the real line.

 

Let $c$ be a point on the real line.

Case I:

If $c<0$, then $f(c)=2 c$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<0$

Case II:

If $c=0$, then $f(c)=f(0)=0$

The left hand limit of $f$ at $x=0$ is,

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}(2 x)=2 \times 0=0$

The right hand limit of $f$ at $x=0$ is,

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{0}}(0)=0$

 

$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

Case IV:

If $c=1$, then $f(c)=f(1)=0$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(0)=0$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x)=4 \times 1=4$

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case $\mathrm{V}$ :

If $c<1$, then $f(c)=4 c$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(4 x)=4 c$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Hence, $f$ is not continuous only at $x=1$

(xii)

The given function $f$ is $f(x)= \begin{cases}\sin x-\cos x, & \text { if } x \neq 0 \\ -1 & \text { if } x=0\end{cases}$

It is evident that $f$ is defined at all points of the real line.

Let $c$ be a real number.

Case I:

If $c \neq 0$, then $f(c)=\sin c-\cos c$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(\sin x-\cos x)=\sin c-\cos c$

 

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, f is continuous at all points x, such that x ≠ 0

Case II:

If $c=0$, then $f(0)=-1$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

Therefore, $f$ is continuous at $x=0$

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

(xiii)

The given function $f$ is $f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$

The given function is defined at all points of the real line.

Let $c$ be a point on the real line.

Case I:

If $c<-1$, then $f(c)=-2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2)=-2$ $\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<-1$

Case II:

If $c=-1$, then $f(c)=f(-1)=-2$

The left hand limit of $f$ at $x=-1$ is,

$\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1}(-2)=-2$

The right hand limit of $f$ at $x=-1$ is,

$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2 \times(-1)=-2$

$\therefore \lim _{x \rightarrow-1} f(x)=f(-1)$

Therefore, $f$ is continuous at $x=-1$

Case III:

If $-1

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c$

 

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points of the interval $(-1,1)$.

Case IV:

If $c=1$, then $f(c)=f(1)=2 \times 1=2$

The left hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(2 x)=2 \times 1=2$

The right hand limit of $f$ at $x=1$ is,

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2=2$

$\therefore \lim _{x \rightarrow 1} f(x)=f(c)$

Therefore, $f$ is continuous at $x=2$

Case V:

If $c>1$, then $f(c)=2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2)=2$

Therefore, $f$ is continuous at all points $x$, such that $x>1$

Thus, from the above observations, it can be concluded that $f$ is continuous at all points of the real line.