Let $f(x)= \begin{cases}\frac{5 e^{1 / x}+2}{3-e^{1 / x}}, & x
eq 0 \ 0, & x=0\end{cases}$ Then at $x=0, x f(x)$ and $f(x)$ are respectively

Question

Let $f(x)= \begin{cases}\frac{5 e^{1 / x}+2}{3-e^{1 / x}}, & x 
eq 0 \ 0, & x=0\end{cases}$ Then at $x=0, x f(x)$ and $f(x)$ are respectively

Aicharya · Accepted Answer

$f(x)=\left\{\begin{array}{cc}\frac{5 e^{1 / x}+2}{3-e^{1 / x}} & x 
eq 0 \ 0 & x=0\end{array}ight.$ Then $x f(x)=\left\{\begin{array}{cc}\frac{x\left(5 e^{1 / x}+2ight)}{3-e^{1 / x}} & x 
eq 0 \ 0 & x=0\end{array}ight.$ Let us check continuity of $x f(x)$ at $x=0$ $\begin{aligned} & 	ext { L.H.L }=\lim _{h ightarrow 0} \frac{(-h)\left[5 e^{-1 / h}+2ight]}{3-e^{-1 / h}}=0 \ & 	ext { R.H.S }=\lim _{h ightarrow 0} \frac{\left[55^{\frac{1}{h}}+2ight]}{3-e^{\frac{1}{h}}}=0 \end{aligned}$ $x f(x)=0$ $	herefore \quad x f(x)$ is continous on $x=0$ Let us check differentiability of $f(x)$ at $x=0$ : $\begin{aligned} & 	ext { L.H.D. }=\lim _{h ightarrow 0} \frac{f(a-h)-\mathrm{f}(0)}{-h} \ & =\lim _{h ightarrow 0} \frac{\frac{5 e^{-1 / n}+2}{3-e^{-1 / n}-0}}{-h} \end{aligned}$ $=\lim _{h ightarrow 0}-\frac{1}{h}\left(\frac{5 e^{-1 / n}+2}{3-e^{-1 / h}}ight) ightarrow$ $\because \quad$ L.H.D is not finite at $x=0$ $	herefore \quad f(x)$ is not differentiable at $x=0$

Let $f(x)= \begin{cases}\frac{5 e^{1 / x}+2}{3-e^{1 / x}}, & x \neq 0 \\ 0, & x=0\end{cases}$ Then at $x=0,…

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