6.56. $\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$.
Solution.

$$
\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)=(\infty-\infty)=\lim _{x \rightarrow 0} \frac{e^{x}-x-1}{x\left(e^{x}-1\right)}=(0 / 0)
$$

Apply L'Hôpital's rule twice:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(e^{x}-x-1\right)^{\prime}}{\left(x\left(e^{x}-1\right)\right)^{\prime}}=\lim _{x \rightarrow 0} \frac{e^{x}-1}{e^{x}-1+x e^{x}}=(0 / 0)= \\
& \quad=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)^{\prime}}{\left(e^{x}-1+x e^{x}\right)^{\prime}}=\lim _{x \rightarrow 0} \frac{e^{x}}{e^{x}(x+2)}=\frac{1}{2}
\end{aligned}
$$