Continuity and limits with transcendental functions: Determine the interval on w

Given
${\displaystyle f\left(x\right)=\frac{e^x}{1-e^x}}$
f(x) is continuous for all points except for the point where denominator is equal to 0.
${\displaystyle 1-e^x=0}$
${\displaystyle e^x=1}$
${\displaystyle {\ln e}^x=\ln 1}$
${\displaystyle x=0}$
Thus interval of continuity are
${\displaystyle (-\mathrm{\infty },0)}$ and ${\displaystyle (0,\mathrm{\infty })}$
${\displaystyle \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{e^x}{1-e^x}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{e^{-h}}{1-e^{-h}}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{1}{\left(\frac{e^h-1}{h}\right)h}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{1}{h}}$
${\displaystyle =\mathrm{\infty }}$
${\displaystyle \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{e^x}{1-e^x}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{e^h}{-(e^h-1)}}$
${\displaystyle =\underset{h\to 0}{lim}\frac{e^h}{-\left(\frac{e^h-1}{h}\right)h}}$
${\displaystyle =\underset{h\to 0}{lim}(-\frac{e^h}{h})}$
${\displaystyle =-\mathrm{\infty }}$