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 ?
 A: 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 $-1/2$. 
A: 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*}
A: You can change it up a bit, by splitting out the denominator to check for form for L'Hospital's Rule:
$$\lim_{x\rightarrow 0} \frac{e^x+\ln \left(\frac{1-x}{e}\right)}{\tan x - x}=
\lim_{x\rightarrow 0} \left(e^x+\ln \left(\frac{1-x}{e}\right)\right)*\lim_{x\rightarrow 0} \frac{1}{\tan x - x}$$
The numerator limit is easy, since it exists:
$$\lim_{x\rightarrow 0} \left(e^x+\ln \left(\frac{1-x}{e}\right)\right)=
\left(e^0 +\ln \left(\frac{1-0}{e} \right)\right)=(1+\ln(e^{-1}))=0$$
The denominator limit is easy, since it exists:
$$\lim_{x\rightarrow 0} {\tan x - x}={\tan 0 - 0}={0}$$
Since you have a valid form, you can apply L'Hospitals Rule. For reference, the rule is:
$$\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f'(x)}{g'(x)}$$
If and only if the new limit exists!
So you have:
$$\lim_{x\rightarrow 0} \frac{e^x+\ln \left(\frac{1-x}{e}\right)}{\tan x - x} =
\lim_{x\rightarrow 0} \frac{e^x+\frac{1}{\frac{1-x}{e}}*\frac{-1}{e}}{\sec^2 x -1}$$
$$=\lim_{x\rightarrow 0} \frac{e^x+\frac{-1}{1-x}}{\sec^2 x -1}$$
This is still $\frac{0}{0}$, so repeat again:
$$=\lim_{x\rightarrow 0} \frac{e^x+(1-x)^{-2}*-1}{-2*\cos^{-3} x *-\sin x}$$
Apply again:
$$=\lim_{x\rightarrow 0} \frac{e^x+-2*(1-x)^{-3}}{6*\cos^{-4} x *\sin^2 x+-2\cos^{-3}x*-\cos x} = 
\lim_{x\rightarrow 0} \frac{e^x+-2*(1-x)^{-3}}{6*\cos^{-4} x *\sin^2 x+2\cos^{-2}x}$$
This limit exists, and evaluates to:
$$\frac{e^0-2*(1-0)^{-3}}{6\cos^{-4} 0 \sin^2 0 +2\cos^{-2} 0}=\frac{1-2}{6/1^4*0^2+2/1^2}=\frac{-1}{2}$$
A: Write $\log{1-x\over e}=\log{1-x}-1$ and use Taylor expansions to order $3$
$$e^x+\log{1-x\over e}=1+x+{x^2\over 2}+{x^3\over 6}-x-{x^2\over 2}-{x^3\over 3}-1+o(x^3)$$
And 
$$\tan x-x={x^3\over 3}+o(x^3)$$
So 
$${e^x+ \ln{1-x\over e }\over \tan x-x}={{-x^3\over 6}+o(x^3)\over {x^3\over 3}+o(x^3)}$$
So the limit we're looking for is $-{1\over 2}$
