Find $\lim\limits_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$ Find $\displaystyle \lim_{x\to0}\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}$
I tried using L'hopital's rule but it got very messy very fast

UPDATE- So reading about the Taylor series this is what I have so far 
$$\lim_{x\to0}\frac{(x-x^3/3!+x^5/5!-\cdots)-(x+x^3/3+2x^5/5+17x^7/315+\cdots)}{(x+x^2/6+3x^5/40+\cdots)-(x-x^3/3+x^5/5-\cdots)}$$
But I'm still stuck

This is a nice exercise which shows that a lot more can be achieved using standard limits than what most beginners would think.
We are going to use the following standard limits $$\lim_{x\to 0} \frac{\sin x} {x} =1,\lim_{x\to 0} \frac{1-\cos x} {x^2}=\frac{1}{2},\lim_{x\to 0}\frac{\arctan x} {x} =1$$ All of these are immediate consequences of the first limit.
The numerator of the given expression can be written as $$\frac{\sin x} {x} \cdot x^3\cdot\frac{\cos x-1}{x^2}\cdot\frac{1}{\cos x} $$ and using the standard limit we can replace the above with $-x^3/2$.
The denominator needs a little more effort. Using the identities $$\arcsin x=\arctan\frac{x} {\sqrt{1-x^2}},\arctan x-\arctan y=\arctan\frac{x-y} {1+xy}$$ we can write the denominator as $$\arctan\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}} $$ and using the limit $\lim_{x\to 0}(\arctan x) /x=1$ the above expression can be replaced by $$\dfrac{\dfrac{x} {\sqrt{1-x^2}}-x } {1+\dfrac{x^2}{\sqrt{1-x^2}}}=\frac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}$$ Thus the desired limit is equal to the limit of $$\dfrac{-x^3/2}{\dfrac{x(1-\sqrt{1-x^2})}{x^2+\sqrt{1-x^2}}}=-\frac{x^2(x^2+\sqrt{1-x^2})}{2(1-\sqrt{1-x^2})}$$ Multiplying numerator and denominator by $1+\sqrt{1-x^2}$ we can simply the expression above as $$-\frac{(x^2+\sqrt{1-x^2})(1+\sqrt {1-x^2})} {2} $$ and this clearly has limit $-1$ as $x\to 0$.
