limit problem - can't get rid of $0$. I am trying to evaluate limit:
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}$$
I tried to use known limit in denominator to get: 
$$\lim_{x\rightarrow 0}\frac{\frac{\arcsin}{x} - \frac{\arctan x}{x}}{x( \frac{e^x-1}{x}\cdot \frac{1}{x}+\frac{1-\cos x}{x^2} -1 -\frac{1}{x})}$$
But I still get $1-1=0$ in numerator. I also tried to use L'Hôpital rule, but it didn't help.
 A: Hint: If you know that $e^x = 1+x+{x^2\over2}+{x^3\over6}+\dots$ and $\cos x=1-{x^2\over2}+{x^4\over24}-\dots$, you can see that the denominator is ${x^3\over6}+\dots$, so the limit reduces to
$$6\lim_{x\to0}{\arcsin x-\arctan x\over x^3}$$
A: Using the Taylor series 
$$f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f^{(3)}(0)+o(x^3)$$
for $f$ is $\arcsin$ and $\arctan$ we find
$$\arcsin x=x+\frac{x^3}{6}+o(x^3)$$
and 
$$\arctan x=x-\frac{x^3}{3}+o(x^3)$$
and we have
$$e^x-\cos x-x^2-x=1+x+\frac{x^2}{2}+\frac{x^3}{6}-1+\frac{x^2}{2}-x^2-x+o(x^3)=\frac{x^3}{6}+o(x^3)$$
hence
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}=\lim_{x\rightarrow 0}\frac{\frac{x^3}{6}+\frac{x^3}{3}+o(x^3)}{\frac{x^3}{6}+o(x^3)}=3$$
A: Just use L‘hospital multiple times:
$$ \lim_{x \rightarrow 0} \frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}
= \lim_{x \rightarrow 0} \frac{\frac{1}{\sqrt{1-x^2}} - \frac{1}{x^2+1}}{e^x+\sin x -2x-1} 
= \lim_{x \rightarrow 0} \frac{\frac{x}{\sqrt{(1-x^2)^3}} + \frac{2x}{(x^2+1)^2}}{e^x+\cos x -2}
= \lim_{x \rightarrow 0} \frac{\frac{3x^2}{\sqrt{(1-x^2)^5}} + \frac{1}{\sqrt{1-x^2)^3}} - \frac{8x^2}{(x^2+1)^3} \frac{2}{(x^2+1)^2}}{e^x-\sin x}
 = \frac{3}{1} =3.
$$
Edit: I edited my pot. My original answer was just wrong.
