Evaluate the limit containing $\arctan{x}$ and $\arcsin{x}$

Here is an approach which relies on trigonometric identities and certain standard limits. The approach involves some amount of labor in algebra and should be used only when more powerful tools like Taylor series or L'Hospital's Rule are forbidden. On the other hand it does show that simple tools can be used to tackle difficult problems if one applies them properly.

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Let $t=\arcsin x$ so that $x=\sin t$ and then $\tan t=x(1-x^2)^{-1/2}$ so that $t=\arctan x(1-x^2)^{-1/2}$ and thus the expression $\arcsin x-\arctan x$ can be written as $$\arctan\frac{x} {\sqrt{1-x^2}}-\arctan x=\arctan\dfrac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}} $$ which can be simplified as $$\arctan\frac{x^3}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})}=\arctan u\text{ (say)} $$ where $u/x^3\to 1/2$ and further $$\frac{2u}{x^3}-1=\frac{1-(1+x^2)\sqrt{1-x^2}}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})}$$ which can be written as $$\frac{-x^2+x^4+x^6}{(x^2+\sqrt{1-x^2})(1+\sqrt{1-x^2})(1+(1+x^2)\sqrt{1-x^2})}$$ and thus $$\lim_{x\to 0}\frac{1}{x^2}\left(\frac{2u}{x^3}-1\right)=-\frac{1}{4}\tag{1}$$ If $L$ is the desired limit in question and $f(x) $ is the function whose limit needs to be evaluated then \begin{align} \log L&=\log\lim_{x\to 0}f(x)=\lim_{x\to 0}\log f(x)\text{ (via continuity of log)} \notag\\ &=\lim_{x\to 0}\frac{2}{x^2}\log\left(\frac{2\arctan u} {x^3}\right)\notag\\ &=\lim_{x\to 0}\frac{2}{x^2}\cdot\dfrac{\log\left(1+\dfrac{2\arctan u-x^3}{x^3}\right)}{\dfrac{2\arctan u-x^3}{x^3}}\cdot\frac{2\arctan u-x^3}{x^3}\notag\\ &=\lim_{x\to 0}\frac{2}{x^2}\cdot\frac{2\arctan u-x^3}{x^3}\notag\\ &=\lim_{x\to 0}\frac{4\arctan u-4u}{x^5}+\frac{4u-2x^3}{x^5}\notag\\ &=\lim_{x\to 0}4\cdot\frac{\arctan u-u} {u^2}\cdot\frac{u^2}{x^6}\cdot x-\frac{1}{2}\text{ (via equation (1))}\notag\\ &=-\frac{1}{2}\notag\end{align} The desired limit $L$ is thus $1/\sqrt{e}$. We have used the limit $$\lim_{x\to 0}\frac{\arctan x-x} {x^2}=\lim_{x\to 0}\frac{x-\tan x} {x^2}=\lim_{x\to 0}\frac{x\cos x-\sin x} {x^2}=0$$ which can be proved without Taylor series or L'Hospital's Rule. Also note that the factor $u^2/x^6=(u/x^3)^2\to (1/2)^2=1/4$.