Question

If $f\left(x\right)=\{\begin{array}{cc}x{e}^{-(\frac{1}{|x|}+\frac{1}{x})};& ifx\ne 0\\ 0;& ifx=0\end{array}$ then which of the following is correct?

• A. $f\left(x\right)$ is continuous and $f^{\prime }\left(0\right)$ does not exist
• B. $f\left(x\right)$ is not continuous
• C. $f\left(x\right)$ is continuous and $f^{\prime \prime }\left(0\right)$ also exists
• D. None of these

Answer 1

Answer: (A)

$f\left(x\right)$ is continuous and $f^{\prime }\left(0\right)$ does not exist

for $f\left(x\right)$ to be continous:

${lim}_{h\to 0}\left(f\right(x+h)=f(x)=f(x-h\left)\right)$

$forx=0$

RHL

${lim}_{h\to 0}h{e}^{-\left(2/h\right)}\to 0$

LHL

${lim}_{h\to 0}-h{e}^{-\left(0\right)}=-h\to 0$

function value at $x=0$ is also zero so function is continuous.

for $f^{\prime }\left(0\right)$
$RHD={lim}_{h\to 0}\frac{f\left(h\right)-f\left(0\right)}{h}={lim}_{h\to 0}{e}^{-2/h}=0$

$LHD={lim}_{h\to 0}\frac{f(-h)-f\left(0\right)}{-h}={lim}_{h\to 0}\frac{-h}{-h}=1$

$LHD\ne RHD$
$f^{\prime }\left(0\right)$ doesn't exist.