How to find the limit of $f(x)$. Let $$f(x) = {\frac{e^{e^x - 1} - \frac{1}{1 - x}}{\ln(\frac{1 + x}{1-x}) - 2\sin x}}$$
I need to find a $ \lim_{x \to 0}f(x)$. As I understand, we have to do some Taylor series decomposition and problem should be pretty easy. But I have had some troubles with it.
P.S : I can decompose $\ln(\frac{1 + x}{1-x})$ as $\ln(1 + \frac{2x}{1-x})$ to use some standard Taylor formulas. I know $\sin x$ Taylor series, but I've got some troubles with $e^{e^x-1}$ and some further transformations.
 A: I think that Taylor series decomposition is good.
But you can also use L'Hôpital's rule.
\begin{align}
&f(x)
=\dfrac{e^{e^x-1}-(1-x)^{-1}}{\ln|1+x|-\ln|1-x|-2\text{sin}x}.\\
\lim_{x\to 0} f(x)
&=\lim_{x\to 0}\dfrac{e^{e^x-1}-(1-x)^{-1}}{\ln|1+x|-\ln|1-x|-2\text{sin}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^x-(1-x)^{-2}}{(1+x)^{-1}+(1-x)^{-1}-2\text{cos}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^{2x}+e^{e^x-1}e^x-2(1-x)^{-3}}{-(1+x)^{-2}+(1-x)^{-2}+2\text{sin}x}\\
&=_{\text{L'Hôpital's rule}}\lim_{x\to 0} \dfrac{e^{e^x-1}e^xe^{2x}+e^{e^x-1}・2e^{2x}+e^{e^x-1}e^xe^x+e^{e^x-1}e^x-6(1-x)^{-4}}{2(1+x)^{-3}+2(1-x)^{-3}+2\text{cos}x}\\
&=\dfrac{1+2+1+1-6}{2+2+2}\\
&=-\dfrac{1}{6}.
\end{align}
A: When you want to use Taylor series of complex expressions, do it for one piece at the time
$$y=e^{e^x-1}\implies \log(y)=e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$
Now
$$y=e^{\log(y)}=1+x+x^2+\frac{5 x^3}{6}+O\left(x^4\right)$$
A: The only problem with Taylor series approach is in finding the series for $e^{e^x-1}$. The series for other terms are available from memory.
Let $t=e^{e^x-1}$ so that $$\frac{dt} {dx} =te^x$$ so that if $$t=a_0+a_1x+a_2x^2+\dots$$ then $a_0=1$ and $$a_1+2a_2x+3a_3x^2+\dots=\left(1+a_1x+a_2x^2+\dots\right)\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \right) $$ Equating coefficients we get $$a_1=1,2a_2=a_1+1,3a_3=\frac{1}{2}+a_1+a_2$$ so that $$a_0=1,a_1=1,a_2=1,a_3=\frac{5}{6}$$ and thus we get $$e^{e^x-1}=1+x+x^2+\frac{5}{6}x^3+o(x^3)$$ Now you can get the desired limit easily as $$\frac{(5/6)-1}{(2/3)-(-1/3)}=-\frac{1}{6}$$ (only the coefficients upto $x^3$ matter in each term and moreover coefficients of $x^0,x^1,x^2$ cancel).
