Question

The function $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{(x+1)}^{2-(\frac{1}{\left|x\right|}+\frac{1}{x})},& x\ne 0\end{array}\\ \begin{array}{cc}0,& x=0\end{array}\end{array}$ is

• A. Discontinuous at only one point
• B. Discontinuous exactly at two points
• C. Continuous everywhere
• D. None of these

Answer 1

The correct option is A Discontinuous at only one point

The only doubtful point is $x=0.$

$L.H.L.=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}{(-h+1)}^{2-(\frac{1}{h}-\frac{1}{h})}=\underset{h\to 0}{lim}{(1-h)}^2=1$

$R.H.L.=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}{(h+1)}^{2-(\frac{1}{h}+\frac{1}{h})}$

$=\underset{h\to 0}{lim}{(1+h)}^{2-\frac{2}{h}}=\underset{h\to 0}{lim}{(1+h)}^2{\left[{(1+h)}^{\frac{1}{h}}\right]}^{-2}$

$=1\times e^{-2}=e^{-2}$

Since $L.H.L\ne R.H.L,$

$\therefore f\left(x\right)$ is not continuous at $x=0.$