Applying L'Hopital's rule for $\lim_{x\to\infty} x((1+1/x)^x-e)$ I know the limit is $-e/2$ but I can't get there. I know I should be using L'Hopitals Rule here, I tried both $0/0$ and $\infty/\infty$, either way it's a big mess. Please help. Maybe you can use the $e^{log(...)}$ trick but I haven't found it to be useful.
Edit: I'm not familiar with the Big-O notation
 A: HINT:  If one wishes to use L'Hospital's Rule
Letting $t=1/x$ we can write 
$$\begin{align}
\lim_{x\to\infty}x\left(\left(1+\frac1x\right)^x-e\right)&=\lim_{t\to 0}\frac1t\left(\left(1+t\right)^{1/t}-e\right)\\\\
&\overbrace{=}^{LHR}\lim_{t\to0}\frac{d(1+t)^{1/t}}{dt}
\end{align}$$
Apply LHR two more times to evaluate the limit of the derivative.
Can you finish?
A: As an alternative to the nice solution with l'Hopital given by Mark Viola, by Taylor's expansion for $t\to 0$


*

*$\log (1+t)=t-\frac12t^2 +o(t^2)$

*$e^t=t+o(t)$
we have
$$\left(1+\frac1x\right)^x=e^{x\log\left(1+\frac1x\right)}=e^{x\left(\frac1x-\frac1{2x^2}+o(1/x^2)\right)}=e^{1-\frac1{2x}+o(1/x)}=e\cdot e^{-\frac1{2x}+o(1/x)}=e\left(1-\frac1{2x}+o(1/x)\right)$$
and therefore
$$x\left(\left(1+\frac1x\right)^x-e\right)=x\left(e-\frac e{2x}+o(1/x)-e\right)=-\frac e{2}+o(1)\to -\frac e 2$$
A: It is easier to do if you let $x=n$ an integer.  $f(n)=(!+\frac{1}{n})^n=\sum_{k=0}^n \binom{n}{k}(\frac{1}{n})^k$.  $e=\sum_{k=0}^\infty\frac{1}{k!}$.  Expand $\binom{n}{k}$ as powers of $n$ (note) and keep the first two terms to get $\frac{1}{k!}(n^k-\frac{k(k-1)n^{k-1}}{2})$.  Use this as an approximation to $f(n)$ and get $n(f(n)-e)\approx -\sum_{k=2}^n\frac{1}{2(k-2)!}\to -\frac{e}{2}$, ignoring higher order terms in the expansion of $e$.
Note:  Dropping the remaining terms is justified since dividing by $n^k$ and multiplying by $n$ leaves powers of $\frac{1}{n}$ which $\to 0$ as $n\to \infty$.
