Answer
$$\frac{1}{2}$$
Work Step by Step
Since
\begin{align*}
\lim _{x \rightarrow 0}\left(\frac{e^{x}}{e^{x}-1}-\frac{1}{x}\right)&=\lim _{x \rightarrow 0}\left(\frac{xe^{x}-e^{x}+1}{x(e^{x}-1)} \right)\\
&=\frac{0}{0}
\end{align*}
Then by using L’Hôpital’s Rule
\begin{align*}
\lim _{x \rightarrow 0}\left(\frac{e^{x}}{e^{x}-1}-\frac{1}{x}\right)&=\lim _{x \rightarrow 0}\left(\frac{xe^{x}-e^{x}+1}{x(e^{x}-1)} \right)\\
&= \lim _{x \rightarrow 0}\left(\frac{xe^{x}+e^x-e^{x}}{xe^{x} +e^x-1 } \right)\\
&= \lim _{x \rightarrow 0}\left(\frac{xe^{x} +e^x}{xe^{x} +2e^x } \right)\\
&=\frac{1}{2}
\end{align*}
