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
 A: You can evaluate this limit using Taylor's expansions. I will write the expansions of the functions below. Note that you can ignore higher powers of x as it is tending to zero. (You will need to think of how many terms to ignore and how many you shouldn't). If you are still not able to evaluate, ask me for help. 
\begin{align}
\sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dotsb \\[4px]
\tan x &= x + \frac{x^3}{3} + \dotsb \\[4px]
\arcsin x &= x +\frac{1^2}{3!}x^3 + \frac{1^23^2}{5!}x^5 + \dotsb \\[4px]
\arctan x &= x - \frac{x^3}{3} + \dotsb 
\end{align}
You can search for more expansions and how to make these expansions in detail on the net or stack exchange itself.

A: 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 $x\to 0$.
A: You were on the right track
$$A=\frac{\sin(x)-\tan(x)}{\arcsin(x)-\arctan(x)}=\frac{\left(x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)\right)-\left(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right) \right)}{\left(x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)\right)-\left(x-\frac{x^3}{3}+\frac{x^5}{5}+O\left(x^7\right) \right)}$$ Simplify to get
$$A=\frac{-\frac{x^3}{2}-\frac{x^5}{8}+O\left(x^7\right)} {\frac{x^3}{2}-\frac{x^5}{8}+O\left(x^7\right) }=-1-\frac{x^2}{2}+O\left(x^4\right)$$ which shows the limit and how it is approached.
