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?
 A: Alternatively, one may use standard Taylor expansions, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}{6}+o(x^4)
\\\tan x&=x+\frac{x^3}{3}+o(x^4)
\\\arctan x&=x-\frac{x^3}{3}+o(x^4)
\\\arcsin x&=x+\frac{x^3}{6}+o(x^4)
\end{align}
$$ giving, as $x \to 0$,
$$ \frac{\arctan x - \arcsin x}{\tan x - \sin x}= \frac{-\frac{x^3}{2}+o(x^4)}{\frac{x^3}{2}+o(x^4)}=-1+o(x) \to -1.
$$
A: 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.
A: Hint:use the L'Hospital's Rule three times
$$\lim\limits_{x\to 0} \frac{(\arctan x - \arcsin x)'''}{(\tan x - \sin x)'''}$$
A: The first step is OK, but you should try and simplify things before going on.
The numerator is
$$
\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}=
\frac{\sqrt{1-x^2}-1-x^2}{(1+x^2)\sqrt{1-x^2}}=
\frac{-x^2(3+x^2)}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)}
$$
The denominator is
$$
\frac{1}{\cos^2x}-\cos x=\frac{1-\cos^3x}{\cos^2x}=
\frac{(1-\cos x)(1+\cos x+\cos^2x)}{\cos^2x}
$$
so you can rewrite your limit as
$$
\lim_{x\to0}
\frac{x^2}{1-\cos x}
\frac{-(3+x^2)\cos^2x}{(1+x^2)\sqrt{1-x^2}(\sqrt{1-x^2}+1+x^2)(1+\cos x+\cos^2x)}
$$
which is easy to compute (the second fraction is not “indeterminate”).
Of course, Taylor expansion is easier.
