Question

If $f\left(x\right)=x\frac{e^{\left[x\right]-\left|x\right|}-2}{\left[x\right]+\left|x\right|}$, then $\underset{x\to 0}{lim}f\left(x\right)$, is

• A. $-1$
• B. $0$
• C. $1$
• D. non-existent

Answer 1

Answer: (D)

non-existent

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{x(e^{-1+x}-2)}{-1-x}[asx\to 0^{-};\left[x\right]=-1;\left|x\right|=-x]$

$=\frac{0(e^{-1}-2)}{-1-0}=0$

${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{x(e^{0-x}-2)}{(0+x)}[asx\to 0^{+};\left[x\right]=0;\left|x\right|=x]$

$={lim}_{x\to 0^{+}}(e^{-x}-2)=1-2=-1$

LHL$\ne$RHL

$\therefore$Limit at $x=0$ does not exist.