How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule, 
$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$
If we use the rule, we will have 
$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{\sec^2x-\cos x}$.
So, I think that I approach this problem in the wrong way.
Have you guy any idea?

Let $\arcsin x = t$ so that $\sin t = x$ and $$\tan t = \dfrac{x}{\sqrt{1 - x^{2}}}$$ so that $$\arcsin x = t = \arctan\dfrac{x}{\sqrt{1 - x^{2}}}$$ and then we can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\tan x - \sin x}\notag\\
&= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin x(1 - \cos x)}\cdot\cos x\notag\\
&= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin x(1 - \cos x)}\notag\\
&= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\sin^{3}x}\cdot(1 + \cos x)\notag\\
&= 2\lim_{x \to 0}\frac{\arctan x - \arcsin x}{x^{3}}\cdot\frac{x^{3}}{\sin^{3}x}\notag\\
&= 2\lim_{x \to 0}\dfrac{\arctan x - \arctan \dfrac{x}{\sqrt{1 - x^{2}}}}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\dfrac{\arctan \left(\dfrac{x - \dfrac{x}{\sqrt{1 - x^{2}}}}{1 + \dfrac{x^{2}}{\sqrt{1 - x^{2}}}}\right)}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\dfrac{\arctan \left(\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}\right)}{\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}}\cdot\dfrac{\dfrac{x(\sqrt{1 - x^{2}} - 1)}{\sqrt{1 - x^{2}} + x^{2}}}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\sqrt{1 - x^{2}} - 1}{x^{2}(\sqrt{1 - x^{2}} + x^{2})}\notag\\
&= 2\lim_{x \to 0}\frac{\sqrt{1 - x^{2}} - 1}{x^{2}}\notag\\
&= 2\lim_{x \to 0}-\frac{1}{\sqrt{1 - x^{2}} + 1}\notag\\
&= -2\cdot\frac{1}{2}\notag\\
&= -1\notag
\end{align}
In the above we have used the standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1 = \lim_{x \to 0}\frac{\arctan x}{x}$$ There is no need to use advanced tools like Taylor's series and L'Hospital's Rule.
