Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \text { then } f(x) \text { is } \\ 0 & , x=0\end{cases}$
Options:
Solution:
2498 Upvotes
Verified Answer
The correct answer is:
continuous for all $x$ but not differentiable at $x=0$
continuous for all $x$ but not differentiable at $x=0$
$f(0)=0 ; f(x)=x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}$
R.H.L. $\lim _{h \rightarrow 0}(0+h) \mathrm{e}^{-2 / h}=\lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^{2 / h}}=0$
L.H.L $\operatorname{Lim}_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0$
Therefore, $f(x)$ is continuous
R.H.D. $\lim _{h \rightarrow 0} \frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{h}=0$
L.H.D. $\operatorname{Lim}_{h \rightarrow 0} \frac{(0-h) \mathrm{e}^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{-h}=1$
Therefore, L.H.D. $\neq$ R.H.D.
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
R.H.L. $\lim _{h \rightarrow 0}(0+h) \mathrm{e}^{-2 / h}=\lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^{2 / h}}=0$
L.H.L $\operatorname{Lim}_{h \rightarrow 0}(0-h) e^{-\left(\frac{1}{h}-\frac{1}{h}\right)}=0$
Therefore, $f(x)$ is continuous
R.H.D. $\lim _{h \rightarrow 0} \frac{(0+h) e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{h}=0$
L.H.D. $\operatorname{Lim}_{h \rightarrow 0} \frac{(0-h) \mathrm{e}^{-\left(\frac{1}{h}-\frac{1}{h}\right)}-h e^{-\left(\frac{1}{h}+\frac{1}{h}\right)}}{-h}=1$
Therefore, L.H.D. $\neq$ R.H.D.
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.