How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$?

Question

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?

Answer 1

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}

Answer 2

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