Limit of $x\left(\left(1 + \frac{1}{x}\right)^x - e\right)$ when $x\to\infty$

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.

You can indeed use l'Hôpital. Consider $$ f(x)=\left(1+\frac{1}{x}\right)^{\!x}=\exp\left(x\log\left(1+\frac{1}{x}\right)\right)=\exp(x\log(x+1)-x\log x) $$ Then $$ f'(x)=f(x)\left(\log(x+1)+\frac{x}{x+1}-\log x-1\right) $$ Then we have to compute $$ \lim_{x\to\infty}\frac{f'(x)}{-1/x^2} $$ Since the limit of $f(x)$ is $e$, we can concentrate on $$ \lim_{x\to\infty}\frac{\log(x+1)-\log x-\dfrac{1}{x+1}}{-1/x^2} $$ and use l'Hôpital again: $$ \lim_{x\to\infty}\frac{\dfrac{1}{x+1}-\dfrac{1}{x}+\dfrac{1}{(x+1)^2}}{2/x^3}= \lim_{x\to\infty}\frac{-x^3}{2x(x+1)^2}=-\frac{1}{2} $$ Reinstating the factor $e$, the limit is indeed $-e/2$.