<s>[INST] 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 ?
 [/INST] This solution uses the standard limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}} &=&\frac{1}{6}
\\
\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}} &=&-\frac{1}{%
3} \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3}.
\end{eqnarray*}
Re-write the original expression as follows
\begin{eqnarray*}
\frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\tan x-x} &=&\frac{e^{x}+\ln
(1-x)-\ln e}{\tan x-x} \\
&=&\frac{\left( e^{x} {\color{red}{-1-x-\frac{1}{2}x^{2}}}\right) + 
{\color{red}{ \left( 1+x+\frac{1}{2}x^{2}\right)}}+\left( \ln (1-x)+ 
\color{blue}{ x+\frac{1}{2}x^{2}}\right) -\color{blue}{ 
(x+\frac{1}{2}x^{2}) }
-1}{\tan x-x} \\
&=&\frac{\left( e^{x}-1-x-\frac{1}{2}x^{2}\right) +\left( \ln (1-x)+x+\frac{1%
}{2}x^{2}\right) }{\tan x-x} \\
&=&\left( \frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1%
}{2}x^{2}}{x^{3}}\right) \left( \frac{x^{3}}{\tan x-x}\right) 
\end{eqnarray*}
passing to the limit one gets
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{e^{x}+\ln \left( \frac{1-x}{e}\right) }{\left(
\tan x-x\right) } &=&\lim_{x\rightarrow 0}\left( \frac{e^{x}-1-x-\frac{1}{2}%
x^{2}}{x^{3}}+\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right) \left( \frac{%
x^{3}}{\tan x-x}\right)  \\
&=&\left( \lim_{x\rightarrow 0}\frac{e^{x}-1-x-\frac{1}{2}x^{2}}{x^{3}}%
+\lim_{x\rightarrow 0}\frac{\ln (1-x)+x+\frac{1}{2}x^{2}}{x^{3}}\right)
\left( \lim_{x\rightarrow 0}\frac{x^{3}}{\tan x-x}\right)  \\
&=&\left( \frac{1}{6}-\frac{1}{3}\right) \left( \frac{3}{1}\right)  \\
&=&-\frac{1}{2}.
\end{eqnarray*}
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