How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$? How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$?
So I think we expand to $x^2$ since the lowest term for $\ln(1+x)$ is $x$
Let $u=\arctan{(x)}$
$\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}=\lim\limits_{x\to 0}\frac{1+u+\frac{u^2}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\lim\limits_{x\to 0}\frac{1+x+o(x^3)+\frac{x^2+o()}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\frac12-\pi$
My answer is rather messy and likely incorrect. Could someone provide an easier way to solve such problem?  
 A: Maybe this way is still convenient, by using L'Hopital's rule twice,
\begin{align}
&\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{[\ln{(1+x)}]^2}\\
&\quad\quad\stackrel{\rm H}{=}\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}\cdot(1+x^2)^{-1}
-e^{\pi x}-\pi xe^{\pi x}}{2\cdot(1+x)^{-1}\cdot\ln{(1+x)}}\\
&\quad\quad\stackrel{\rm H}{=}\frac{1}{2}\lim\limits_{x\to 0}
\frac
{e^{\arctan{(x)}}\cdot(1+x^2)^{-2}-2x\cdot e^{\arctan{(x)}}\cdot(1+x^2)^{-2}
-2\pi e^{\pi x}-\pi^2 xe^{\pi x}}
{-(1+x)^{-2}\cdot\ln{(1+x)}+(1+x)^{-2}}\\
&\quad\quad=\frac{1}{2}\cdot\left(\frac{1-0-2\pi-0}{0+1}\right)\\
&\quad\quad=\frac{1}{2}-\pi.
\end{align}
A: Let's try to simplify the expression first. We have
\begin{align}
L &= \lim_{x \to 0}\frac{e^{\arctan x} - xe^{\pi x} - 1}{(\log(1 + x))^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - xe^{\pi x} - 1}{\left(\dfrac{\log(1 + x)}{x}\right)^{2}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x + \arctan x - xe^{\pi x} - 1}{1^{2}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x  - 1}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x  - 1}{(\arctan x)^{2}}\cdot\frac{(\arctan x)^{2}}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x  - 1}{(\arctan x)^{2}} + \lim_{x \to 0}\frac{\arctan x - x + x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{t \to 0}\dfrac{e^{t} - t  - 1}{t^{2}} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} + \lim_{x \to 0}\frac{1 - e^{\pi x}}{x}\text{ (putting }t = \arctan x)\notag\\
&= \lim_{t \to 0}\dfrac{e^{t} - 1}{2t} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} - \lim_{x \to 0}\frac{e^{\pi x} - 1}{\pi x}\cdot \pi \text{ (using LHR for first limit)}\notag\\
&= \frac{1}{2} - \pi -  \lim_{t \to 0}\frac{t - \tan t}{\tan^{2} t}\text{ (putting }t = \arctan x)\notag\\
&= \frac{1}{2} - \pi -  \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\cdot\frac{t^{2}}{\tan^{2}t}\notag\\
&= \frac{1}{2} - \pi -  \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\notag
\end{align}
The last limit for $(t - \tan t)/t^{2}$ as $t \to 0$ can be easily shown to be $0$ using the inequalities $$\sin t < t < \tan t$$ for $0 < t < \pi/2$. Hence the desired limit is $\dfrac{1}{2} - \pi$.
