a limit problem? $$\lim_{x\to 0}{e^x+ \ln{1-x\over e }\over \tan x-x} $$
i tried doing l'hospital. but couldn't do ! could anyone help ?

This screams "Taylor" to me. First, $\log\frac{1-x}e=\log (1-x)-\log e=\log(1-x)-1$. We have
$$
e^x=1+x+\frac{x^2}2+\frac{x^3}6+o(x^4),\ \ \log(1-x)=-x+\frac{x^2}2-\frac{x^3}3+o(x^4),
$$
$$\tan x=x-\frac{x^3}3+o(x^5).
$$
So
\begin{align}
{e^x+ \ln{1-x\over e }\over \tan x-x}&=\frac{1+x+\frac{x^2}2+\frac{x^3}6+o(x^4)+(-x+\frac{x^2}2-\frac{x^3}3+o(x^4))-1}{x-\frac{x^3}3+o(x^5)-x}\\
&=\frac{\frac{x^3}6+o(x^4)}{-\frac{x^3}3+o(x^5)}
=\frac{\frac12+o(x)}{-1+o(x)},
\end{align}
so the limit as $x\to0$ is $-1/2$. 
