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

Question:

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

(i) $f(x)=\left\{\begin{array}{cc}x^{3}-x^{2}+2 x-2, & \text { if } x \neq 1 \\ 4 & \text { if } x=1\end{array}\right.$

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

(iii) $f(x)= \begin{cases}\frac{\sin x}{x}, & \text { if } x<0 \\ 2 x+3, & x \geq 0\end{cases}$

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

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

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

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

(viii) $f(x)= \begin{cases}|x-3|, & \text { if } x \geq 1 \\ \frac{x^{2}}{4}-\frac{3 x}{2}+\frac{13}{4}, & \text { if } x<1\end{cases}$

(ix) $f(x)=\left\{\begin{array}{lc}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-33\end{array}\right.$

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

(xi) $f(x)=\left\{\begin{array}{llc}2 x, & \text { if } & x<0 \\ 0, & \text { if } & 0 \leq x \leq 1 \\ 4 x, & \text { if } & x>1\end{array}\right.$

(xii) $f(x)=\left\{\begin{array}{cc}\sin x-\cos x, & \text { if } x \neq 0 \\ -1, & \text { if } x=0\end{array}\right.$

(xiii) $f(x)=\left\{\begin{array}{llc}-2, & \text { if } & x \leq-1 \\ 2 x, & \text { if } & -1 Solution: (i) When$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$It is observed that the left and right hand limits of f at x = 1 do not coincide. Thereforef is not continuous at x = 1 Case$\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)$Thereforef is continuous at all points x, such that x 0 Case II: If$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.