Question

If If $f(x)=\{\begin{array}{ll}\frac{e^{[x]+|x|}-1}{[x]+|x|}& :x\ne 0\\ -1& :x=0\end{array}$ (where $\left[.\right]$ denotes the greatest integer function), then

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $\underset{x\to 0^{+}}{\lim }f\left(x\right)=-1$
• C. $\underset{x\to 0^{-}}{\lim }f\left(x\right)=1$
• D. $\underset{x\to 0^{+}}{\lim }f\left(x\right)=1$

Answer 1

Answer: (D)

$f(x)=\{\begin{array}{cl}\frac{e^{[x]+|x|}-1}{[x]+|x|}& x\ne 0\\ -1& x=0\end{array}$
$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{e^{\left[x\right]+|x|}-1}{\left[x\right]+|x|}=\frac{e^{-1}-1}{-1}=\frac{e-1}{e}$
$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{e^{\left[x\right]+|x|}-1}{\left[x\right]+|x|}$ $=\underset{x\to 0^{+}}{\lim }\frac{e^x-1}{x}=1$
$\because LHL\ne RHL$ at $x=0\Rightarrow f\left(x\right)$ is discontinuous at $x=0$