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
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?
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
$$x\left(\left(1+\frac1x\right)^x-e\right)=x\left(e-\frac e{2x}+o(1/x)-e\right)=-\frac e{2}+o(1)\to -\frac e 2$$
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 $n\to \infty$.
