limit as x goes to zero $$\lim_{x \to 0} \frac{x^2\sin(\frac{1}{x})+x}{(1+x)^\frac{1}{x} -e} =  $$ 
Can anyone help with this limit? I know I have to split it up and apply L'hopitals rule, and the answer is $$\frac{-2}{e}$$ I just cant get it to work out on paper.
 A: hint
The numerator can be set as
$$x( x\sin(\frac 1x) +1 )$$
observe that $$\lim_{x\to 0}\Bigl(x\sin(\frac 1x) +1 \Bigr)  =1$$
now, we need to compute $$\lim_{x\to 0}\frac{x}{(1+x)^\frac 1x -e}$$
with $$(1+x)^\frac 1x=e^{\frac 1x\ln(1+x)}=e^{f(x)}.$$
now use l'Hospital to get
$$\lim_{x\to 0}\frac{1}{f'(x)e^{f(x)}}$$
where $$f'(x)=\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}$$
A: Interesting limit. After the simplification already proposed you get to
$$\mathcal L =\lim_{x\to 0}\frac{x}{(1+x)^{\frac1{x}}-e},$$
which, by the way, could be interpreted as the inverse of the derivative in $0$ of the function
$$f(x)=\begin{cases}(1+x)^{\frac1{x}} & (x>-1 \land x\neq 0)\\e& (x=0).\end{cases}$$
I would go like that
\begin{eqnarray}
\mathcal L &=& \lim_{x\to 0}\frac{x}{e^{\frac1{x}\log(1+x)}-e}=\\
&=&\lim_{x\to 0}\frac{x}{e\left[e^{\frac1{x}\log(1+x)-1}-1\right]}.
\end{eqnarray}
Now use 
$$\frac{\log(1+\alpha(x))}{\alpha(x)} \to 1$$
and
$$\frac{e^{\alpha(x)}-1}{\alpha(x)}\to 1$$
when $\alpha(x) \to 0$, to get
\begin{eqnarray}
\mathcal L &=& \frac1e\cdot\lim_{x\to 0}\frac{x}{\frac1{x}\log(1+x)-1}=\\
&=&\frac1e\cdot\lim_{x\to 0} \frac{x^2}{\log(1+x)-x}.
\end{eqnarray}
Use now MacLaurin or de l'Hospital to handle it.
A: The exercise is meant to make you not use l'Hôpital, because the derivative of the numerator has no limit at $0$. It's not difficult to do it with Taylor expansion.
You know that
$$
\frac{1}{x}\log(1+x)=\frac{1}{x}(x-x^2/2+o(x^2))=1-x/2+o(x)
$$
so
$$
(1+x)^{1/x}-e=e^{1-x/2+o(x)}-e=e\cdot e^{-x/2+o(x)}-e=e(1-x/2+o(x))-e=-ex/2+o(x)
$$
Note also that
$$
x^2\sin\frac{1}{x}=o(x)
$$
and therefore the limit is
$$
\lim_{x\to0}\frac{x+o(x)}{-ex/2+o(x)}=-\frac{2}{e}
$$
