One method is via the substitution $x = 1/y$. As $x \to \infty$, $y \to 0+$.
The given function can be written as
$$
\begin{eqnarray*}
\frac 1 y \left[ (1+y)^{1/y} - {\mathrm e} \right]
&=&
\frac{1}{y} \left[ \exp\left(\frac{\ln (1+y)}{y} \right) - \mathrm e \right]
\\&=&
\frac{\mathrm e}{y} \left[ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 \right]
\\ &=&
\mathrm e \cdot \frac{\ln(1+y)-y}{y^2}
\cdot \frac{ \exp\left(\frac{\ln (1+y) - y}{y} \right) - 1 }{\frac{\ln (1+y) - y}{y}}
\end{eqnarray*}
$$
Can you go from here? The last factor approaches $1$ by the standard limit $\frac{{\mathrm e}^z -1}{z} \to 1$ as $z \to 0$. The limit of the middle factor can be evaluated by L'Hôpital's rule.
The statement of l'Hopital's rule found in Rudin's Principles of Mathematical Analysis (page 109) is:
$5.13$ $\,\,$ Theorem $\,\,\,\,\,\,\,$ Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty\leq a < b\leq +\infty$. Suppose
$$
\frac{f'(x)}{g'(x)}\to A\,\textrm{ as }\,x\to a.
$$
If
$f(x)\to 0$ and $g(x)\to 0$ as $x\to a$, or if $g(x)\to +\infty$ as $x\to a$, then
$$
\frac{f(x)}{g(x)}\to A\,\textrm{ as }\,x\to a.
$$
The analogous statement is of course also true if $x\to b$, or if $g(x)\to-\infty$ [...]. Let us note that we now use the limit concept in the extended sense of Definition $4.33$.
It seems that what $A$ is supposed to be is a little ambiguous. Real? Possibly infinite? However, we can resolve this enigma with a quick look at definition $4.33$:
$4.33$ $\,\,$ Definition $\,\,\,\,\,\,\,$ Let $f$ be a real function defined on $E\subset \Bbb{R}$. We say that
$$
f(t)\to A\,\textrm{ as }\,t\to x,
$$
where $A$ and $x$ are in the extended real number system, if for every neighborhood $U$ of $A$ there is a neighborhood $V$ of $x$ such that $V\cap E$ is not empty, and such that $f(t)\in U$ for all $t\in V\cap E$, $t\neq x$.
So indeed, $A$ is allowed to be infinite (it's in the extended real number system)! We also see that Rudin treats the cases of $A = \pm\infty$ in the proof of l'Hopital's rule, so if the limit of the quotient of derivatives is infinite, the original limit must have been also. That is, there is no example (let alone a simple one) where the limit of the quotient of derivatives is infinite, but the original limit is finite.
Best Answer
I don't know where this comes from so I can't say if there's a mistake or not. But it's no coincidence. L'Hospital's rule is applicable here. The version of the rule as stated in most textbooks says you need to have a limit of the form $$\frac{\infty}{\infty}$$ A more general version of the rule applies to limits of the form $$\frac{\text{anything}}{\infty}$$ There's not even a requirement that the limit of the numerator exist! Of course, all other (often neglected) hypotheses of the rule still have to be satisfied.