How do I evaluate the limit $\lim_{n\to\infty}n((1+1/n)^n-e)$? Could anyone show me how to evaluate this limit?
$$
\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)
$$
Thanks!
 A: Since $\log(1+x)=x-x^2/2+O(x^3)$, we get
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=n\left(\frac1n-\frac1{2n^2}+O\left(\frac1{n^3}\right)\right)\\
&=1-\frac1{2n}+O\left(\frac1{n^2}\right)
\end{align}
$$
Therefore, since $e^x=1+x+O\left(x^2\right)$, we have
$$
\left(1+\frac1n\right)^n=e\left(1-\frac1{2n}+O\left(\frac1{n^2}\right)\right)
$$
From here, it is easy to see that
$$
n\left(\left(1+\frac1n\right)^n-e\right)=-\frac{e}{2}+O\left(\frac1n\right)
$$
Thus,
$$
\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)=-\frac{e}{2}
$$
A: Let's see another way 
$$\lim_{n\to\infty}n\left(\left(1+\frac1n\right)^n-e\right)=$$
$$\lim_{n\to\infty}en\left(\frac{\left(1+\frac1n\right)^n}{e}-1\right)=$$
$$\lim_{n\to\infty}en\ln\left(\frac{\left(1+\frac1n\right)^n}{e}\right)\times \lim_{n\to\infty}\frac{\left(\displaystyle\frac{\left(1+\frac1n\right)^n}{e}-1\right)}{\ln\left(\displaystyle\frac{\left(1+\frac1n\right)^n}{e}\right)}=$$
$$\lim_{n\to\infty}en\ln\left(\frac{\left(1+\frac1n\right)^n}{e}\right)=  $$
$$\lim_{n\to\infty}e \left(n^2\ln\left(1+\frac{1}{n}\right)-n\right)=-\frac{e}{2} $$
where I used $\displaystyle \lim_{x\to1} \frac{x-1}{\ln x}=1$, and then
$$\lim_{x\to\infty} \left(x^2\ln\left(1+\frac{1}{x}\right)-x\right)= $$
$$\lim_{y\to0} \left(\frac{1}{y^2}\ln\left(1+y\right)-\frac{1}{y}\right)= $$
$$\lim_{y\to0} \left(\frac{\ln(1+y)-y}{y^2}\right)= $$
that by l'Hôpital's rule turns into
$$ \lim_{y\to0} -\frac{1}{2(y+1)}=-\frac{1}{2}$$
Chris.
A: 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,
$$A = e \lim_{n\rightarrow\infty} \frac{\left(\frac{-1/n^2}{1+1/n} + \frac{1}{(n+1)^2}\right)}{2/n^3}
= e \lim_{n\rightarrow\infty} \frac{-n^2}{2(n+1)^2}
= -\frac{e}{2}
$$
