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.

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
$$\lim_{x\rightarrow 0}\frac{\arcsin x - \arctan x}{e^x-\cos x -x^2 -x}=\lim_{x\rightarrow 0}\frac{\frac{x^3}{6}+\frac{x^3}{3}+o(x^3)}{\frac{x^3}{6}+o(x^3)}=3$$
