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

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