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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?

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    $\begingroup$ Your answer is correct! $\endgroup$ – Mark Viola Apr 20 '16 at 4:24
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    $\begingroup$ As you can see, your answer is not only correct but also easier than the two answers you have received so far. In general, don't be afraid to use Taylor expansion as it is not only systematic but also usually the fastest! $\endgroup$ – user21820 Apr 20 '16 at 8:43
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    $\begingroup$ Of course, you ought to fill in your empty little-o expressions, but I'm sure you do know how to do that. $\endgroup$ – user21820 Apr 20 '16 at 8:45
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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}

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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 $\dfrac{1}{2} - \pi$.

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