<s>[INST] 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
 [/INST] 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.
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