Find the limit of $\lim\limits_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ Find the limit of $\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ approaches right of zero.
The answer is $\frac{1}{2}$ but I keep getting 1. Here's what I have:
Since $\lim_{x\to0^+} \frac{\ln (1+x)}{x}$ is of indeterminate form (0/0), we can apply L Hôpitals.
So now I have 
$\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{1}{1+x}}{x}$, which is also of indeterminate form (0/0).
So by L Hôpitals, $\lim_{x\to0^+}\frac{1}{e} \frac{{- e^ \frac{1}{1+x} \frac{-1}{(1+x)^2}}}{x}$ which is equal to 1.
But as I said, the answer is $\frac12$. Can you tell me where did I go wrong? Thanks!

$$\lim_{x\rightarrow0}\frac{e- e^ \frac{\ln (1+x)}{x}}{ex}=-\lim_{x\rightarrow0}\frac{e^ \frac{\ln (1+x)}{x}\left(\frac{x}{1+x}-\ln(1+x)\right)}{ex^2}=-\lim_{x\rightarrow0}\frac{x-(1+x)\ln(1+x)}{x^2}=$$
$$=-\lim_{x\rightarrow0}\frac{1-\ln(1+x)-1}{2x}=\frac{1}{2}.$$
