Answer
$$
\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) =\frac{1}{2}
$$
Work Step by Step
$$
\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)
$$
First notice that
$$
\lim _{x \rightarrow 0^{+}} \frac{1}{x} \,\,\, \rightarrow \infty
$$
and
$$
\lim _{x \rightarrow 0^{+}} \frac{1}{e^{x}-1} \,\,\, \rightarrow \infty ,
$$
so the limit is indeterminate and has the form $\infty -\infty $. Here we use a common denominator:
$$\begin{aligned}
\lim _{x \rightarrow 0^{+}}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) &=\lim _{x \rightarrow 0^{+}} \frac{e^{x}-1-x}{x\left(e^{x}-1\right)} \rightarrow \frac{0}{0} \\
& \stackrel{\text { H }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}-1}{x e^{x}+e^{x}-1} \rightarrow \frac{0}{0} \\
& \stackrel{\text { H }}{=} \lim _{x \rightarrow 0^{+}} \frac{e^{x}}{x e^{x}+e^{x}+e^{x}} \\
& =\frac{1}{0+1+1} \\
&=\frac{1}{2}
\end{aligned}
$$
Note that the use of l’Hospital’s Rule is justified because:
$ e^{x}-1-x \rightarrow 0 \quad $ and $ \quad x\left(e^{x}-1\right) \rightarrow 0 $ as$ \rightarrow 0^{+}$
also,
$ e^{x}-1 \rightarrow 0 \quad $ and $ \quad x e^{x}+e^{x}-1 \rightarrow 0 $ as$ \rightarrow 0^{+}$.