Question

Find if the given below function is continuous or discontinuous at the indicated point
$f\left(x\right)=\{\begin{array}{cc}{e}^{\frac{1}{x}}ifx\ne 0& \\ 1+e^{\frac{1}{x}}ifx=0& \\ 0,& \end{array}$
at $x=0$

Answer 1

Find left hand limit at $x=0$
$f\left(x\right)=\{\begin{array}{cc}{e}^{\frac{1}{x}}ifx\ne 0& \\ 1+e^{\frac{1}{x}}ifx=0& \\ 0,& \end{array}$
at $x=0$
$L.H.L={lim}_{x\to 0-}f\left(x\right)={lim}_{h\to 0+}f(0-h)$
${lim}_{h\to 0+}\frac{e^{\left(\frac{1}{0-h}\right)}}{1+e^{\left(\frac{1}{0-h}\right)}}$
$={lim}_{h\to 0+}\frac{e^{\left(\frac{1}{-h}\right)}}{1+e^{\left(\frac{1}{-h}\right)}}$
$=\frac{e^{-\infty }}{1+e^{-\infty }}$=$\frac{0}{1+0}=0$
Find right hand limit at $x=0$
$R.H.L={lim}_{x\to 0+}f\left(x\right)={lim}_{h\to 0+}f(0+h)$
$={lim}_{h\to 0+}\frac{e^{\left(\frac{1}{0+h}\right)}}{1+e^{\frac{1}{0+h}}}$
${lim}_{h\to 0+}\frac{1}{e^{\left(\frac{-1}{h}\right)}+1}$
$\frac{1}{e^{-\infty }+1}$
$\frac{1}{0+1}{e}^{-\infty }=0$
$=\mathrm{1......}\left(3\right)$
Thus $R.H.L\ne L.H.L\text{at}x=$