TITLE: How do I evaluate the limit $\lim_{n\to\infty}n((1+1/n)^n-e)$?
QUESTION [5 upvotes]: Could anyone show me how to evaluate this limit?
$$
\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)
$$
Thanks!

REPLY [4 votes]: It's possible to do with L'Hopital too but the derivation is a bit lengthy:
$$A = \lim_{n\rightarrow\infty} \frac{\left(1+1/n\right)^n - e}{1/n}$$
Having $0/0$ we can try to use L'Hopital.
Because
$$ \frac{d}{dn} \left(1+1/n\right)^n
= \frac{d}{dn} \exp\left(n \ln\left(1+1/n\right)\right)
= \left(1+1/n\right)^n \left(\ln\left(1+1/n\right) + \frac{n\left(-1/n^2\right)}{1+1/n}\right)
$$
we have
$$A = \lim_{n\rightarrow\infty} \frac{\left(1+1/n\right)^n \left(\ln\left(1+1/n\right) - \frac{1}{n+1}\right)}{-1/n^2}
= e \lim_{n\rightarrow\infty} \frac{\left(\ln\left(1+1/n\right) - \frac{1}{n+1}\right)}{-1/n^2}$$
Which is $0/0$ again (Because $\ln(1+\epsilon)\approx \epsilon$ for $\epsilon\ll 1$) , using L'Hopital once more,
$$A = e \lim_{n\rightarrow\infty} \frac{\left(\frac{-1/n^2}{1+1/n} + \frac{1}{(n+1)^2}\right)}{2/n^3}
= e \lim_{n\rightarrow\infty} \frac{-n^2}{2(n+1)^2}
= -\frac{e}{2}
$$