Question

Find if limit of function exists as x tends to zero

$f\left(x\right)=\{\begin{array}{c}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}x\ne 0\\ 0x=0\end{array}$

Answer 1

According to,

$\begin{array}{c}f\left(x\right)=\{\frac{e^{\frac{1}{x}}-1}{\begin{array}{c}{e}^{\frac{1}{x}}+1\\ 0\end{array}}\begin{array}{c}x\ne 0\\ x=0\end{array}\\ LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{-}}\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}=\frac{0-1}{0+1}=-1\\ \\ RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0^{+}}\frac{e^{\frac{1}{x}}(1-e^{-\frac{1}{x}})}{e^{\frac{1}{x}}(1+e^{-\frac{1}{x}})}=\frac{1-0}{1+0}=1\\ \therefore LHL\ne RHL\\ {lim}_{x\to 0}f\left(x\right)doesnotexist.\end{array}$