Limit of $x\left(\left(1 + \frac{1}{x}\right)^x - e\right)$ when $x\to\infty$ I am stuck on how to calculate the following limit:
$$\lim_{x\to\infty}x\left(\left(1 + \frac{1}{x}\right)^x - e\right).$$
Definitely, it has to be through l'Hôpital's rule. We know that $\lim_{x\to\infty} (1+(1/x))^x = e$. So, I wrote the above expression as
$$\lim_{x\to\infty}\frac{\left(1 + \frac{1}{x}\right)^x - e}{1/x}.$$Both numerator and denominator tend to zero as x tends to infinity. I applied l'Hôpital's rule twice, but I got the limit equal to infinity which is clearly wrong. In the book, it says that the limit should be $-e/2$.
Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your l'Hôpital's rule to in both cases? Thanks a lot
Also, guys can you tell me how to write equations in this forum? Thanks.
 A: 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 $-e/2$.
A: 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. 
A: I'll use the rule twice where you see a $\ast$, but this probably isn't the most elegant treatment:$$\begin{align}\lim_{x\to\infty}\frac{(1+1/x)^x-e}{1/x}&=\lim_{y\to0}\frac{(1+y)^{1/y}-e}{y}\\&\stackrel{\ast}{=}\lim_{y\to0}\frac{(1+y)^{1/y}\frac{d}{dy}\frac{\ln(1+y)}{y}}{1}\\&=\lim_{y\to0}(1+y)^{1/y}\frac{1-(1+y)\frac1y\ln(1+y)}{y(1+y)}\\&=e\lim_{y\to0}\frac{1-(1+y)\frac1y\ln(1+y)}{y(1+y)}\\&\stackrel{\ast}{=}e\lim_{y\to0}\frac{(\ln(1+y)-y)/y^2}{1+2y}\\&=e\frac{-1/2}{1+0}=-\frac{e}{2}.\end{align}$$Another solution that doesn't use the rule notes that$$(1+1/x)^x=\exp x\ln(1+\frac1x)=e\cdot\exp\left(-\frac{1}{2x}+o\left(\frac1x\right)\right)=e\left(1-\frac{1}{2x}+o\left(\frac1x\right)\right).$$Thus$$x((1+1/x)^x-e)=-\frac{e}{2}+o(1).$$
A: It's simpler to use Taylor formula at order $2$. Indeed, set $u=\frac1x$. The expression becomes
$$x\biggl(\Bigl(1+\frac{1}{x}\Bigr)^x-\mathrm e\biggr)=\frac{(1+u)^{\tfrac1u}-\mathrm e}u=\frac{\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e}u.$$
Now the numerator is
\begin{align}
\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e&= \mathrm e^{\tfrac{u-\frac12 u^2 +o(u^2)}u}-\mathrm e=\mathrm e^{1-\frac{1}2u+o(u)}-\mathrm e=
\mathrm e\bigl(\mathrm e^{\tfrac{u}2+o(u)}-1\bigr)\\
&=\mathrm e\Bigl(1+\frac u2+o(u)-1\Bigr)=\frac{\mathrm e\mkern2mu u}2+o(u)\sim_0 \frac{\mathrm e\mkern2mu u}2
\end{align}
so that $$\frac{\mathrm e^{\tfrac{\ln(1+u)}u}-\mathrm e}u\sim_0 \frac{\mathrm e\mkern2mu u}{2u}=\frac{\mathrm e}2.$$
