Question

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

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

Answer 1

Answer: (D)

$f(x)=\left\{\begin{array}{cl}\frac{e^{[x]+|x|}-1}{[x]+|x|} & x \neq 0 \\ -1 & x=0\end{array}\right.$
$\underset{x \rightarrow 0^{-}}{l i m}f\left(x\right)=\underset{x \rightarrow 0^{-}}{l i m}\frac{e^{\left[x\right] + \left|x\right|} - 1}{\left[x\right] + \left|x\right|}=\frac{e^{- 1} - 1}{- 1}=\frac{e - 1}{e}$
$\underset{x \rightarrow 0^{+}}{l i m}f\left(x\right)=\underset{x \rightarrow 0^{+}}{l i m}\frac{e^{\left[x\right] + \left|x\right|} - 1}{\left[x\right] + \left|x\right|}$ $=\underset{x \rightarrow 0^{+}}{l i m}\frac{e^{x} - 1}{x}=1$
$\because LHL\neq RHL$ at $x=0\Rightarrow f\left(x\right)$ is discontinuous at $x=0$