Question

Discuss the continuity of the following functions at the indicated point(s):
(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Answer 1

(i) Given:
$f\left(x\right)=\left\{\begin{array}{l}\left|x\right|\cos \left(\frac{1}{x}\right),x\ne 0\\ 0,x=0\end{array}\right.$
We observe
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left|x\right|\cos \left(\frac{1}{x}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left|x\right|\underset{x\to 0}{\lim }\cos \left(\frac{1}{x}\right) \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=0\times \underset{x\to 0}{\lim }\cos \left(\frac{1}{x}\right)=0 \end{gathered}$$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(ii) Given:
$f\left(x\right)=\left\{\begin{array}{l}{x}^2\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right.$
We observe
$\underset{x\to 0}{\lim }{x}^2\sin \left(\frac{1}{x}\right)=\underset{x\to 0}{\lim }{x}^2\underset{x\to 0}{\lim }\sin \left(\frac{1}{x}\right)=0\times \underset{x\to 0}{\lim }\sin \left(\frac{1}{x}\right)=0$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$
Hence, f(x) is continuous at x = 0.

(iii) Given:
$f\left(x\right)=\left\{\begin{array}{l}\left(x-a\right)\sin \left(\frac{1}{x-a}\right),x\ne a\\ 0,x=a\end{array}\right.$

Putting x−a = y, we get
$\underset{x\to a}{\lim }\left(x-a\right)\sin \left(\frac{1}{x-a}\right)=\underset{y\to 0}{\lim }y\sin \left(\frac{1}{y}\right)$$=\underset{y\to 0}{\lim }y\underset{y\to 0}{\lim }\sin \left(\frac{1}{y}\right)=0\times \underset{y\to 0}{\lim }\sin \left(\frac{1}{y}\right)=0$
$\Rightarrow \underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)=0$
Hence, f(x) is continuous at x = a.

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

We observe
$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{e^x-1}{\log \left(1+2x\right)} \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{e^x-1}{{\displaystyle \frac{2x\log \left(1+2x\right)}{2x}}} \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{1}{2}\underset{x\to 0}{\lim }\frac{\left({\displaystyle \frac{e^x-1}{x}}\right){\displaystyle }}{{\displaystyle \left(\frac{\log \left(1+2x\right)}{2x}\right)}} \\ \Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{1}{2}\times \frac{\left(\underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{x}}\right)}{\left(\underset{x\to 0}{\lim }{\displaystyle \frac{\log \left(1+2x\right)}{2x}}\right)}=\frac{1}{2}\times \frac{1}{1}=\frac{1}{2} \end{gathered}$$
And, $f\left(0\right)=7$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)\ne f\left(0\right)$

Hence, f(x) is discontinuous at x = 0.

(v) Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{1-x^n}{1-x},x\ne 1\\ n-1,x=1\end{array}\right.$

Here, $f\left(1\right)=n-1$

$$\begin{gathered} \underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\frac{1-x^n}{1-x} \\ \Rightarrow \underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\left[{\left(1-x\right)}^{n-1}+{}^{n}C_1{\left(1-x\right)}^{n-2}x+{}^{n}C_2{\left(1-x\right)}^{n-3}{x}^2+...+{}^{n}C_{n-1}{\left(1-x\right)}^0{x}^{n-1}\right] \end{gathered}$$
$\Rightarrow \underset{x\to 1}{\lim }f\left(x\right)=0+0...+{\left(1\right)}^{n-1}=1\ne f\left(1\right)$

Thus, $f\left(x\right)\mathrm{is}\mathrm{discontinuous}\mathrm{at}x=1$.

(vi) Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{\left|x^2-1\right|}{x-1},x\ne 1\\ 2,x=1\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}x+1,x<-1\\ \begin{array}{c}-x-1,-1\le x<1\\ x+1,x>1\end{array}\\ 2,x=1\end{array}\right.$
We observe
(LHL at x = 1) = $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }-\left(1-h\right)-1=\underset{h\to 0}{\lim }-2+h=-2$
And, $f\left(1\right)=2$

$\Rightarrow \underset{x\to 1^{-}}{\lim }f\left(x\right)\ne f\left(1\right)$

Hence, f(x) is discontinuous at x = 1.

(vii) Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{2\left|x\right|+x^2}{x},x\ne 0\\ 0,x=0\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{2x+x^2}{x},x>0\\ \frac{-2x+x^2}{x},x<0\\ 0,x=0\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\left(x+2\right),x>0\\ \left(x-2\right),x<0\\ 0,x=0\end{array}\right.$

We observe

(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(-h\right)=\underset{h\to 0}{\lim }\left[-h-2\right]=-2$
(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(h\right)=\underset{h\to 0}{\lim }\left(2+\mathrm{h}\right)=2$

$\Rightarrow \underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$

Hence, f(x) is discontinuous at x = 0.