TITLE: Limit of $x\left(\left(1 + \frac{1}{x}\right)^x - e\right)$ when $x\to\infty$
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: 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 $-e/2$.