limit with Taylor expansion

Question

I have a limit:

$$\ \lim_{x \to 0} = \frac{2\arcsin(1-\cos(x))-\tan^2x}{\ln(1+x^2)-\sin^2(x)}$$

I have substituted all the functions with Taylor series and divided by $\ x^6 $

$$\ \frac{2(\frac{1}{2x^3}+\frac{o(x^2)}{x^2}*\frac{1}{x^3}+\frac{(\frac{1}{2}+\frac{o(x^2)}{x^2})^3}{6}+\frac{o(x^3)}{x^3}*\frac{1}{x^2})-(\frac{1}{x}+\frac{1}{3}+\frac{o(x^3)}{x^3})^2}{\frac{1}{x^3}+\frac{o(x^2)}{x^2}*\frac{1}{x^3}-(\frac{1}{x^2}-\frac{1}{6}+\frac{o(x^3)}{x^3})^2} $$

Is it correct? Additionally, I am not sure, how can I calculate the limit for $\ x\to 0.$ For $\ \infty$ it seems to be not to scary.

EDIT
I use: $$\ \ln(1+x^2)=x^2+o(x^2)\\ \sin(x)=x-\frac{x^3}{6}+o(x^3) \\ \cos(x)=1-\frac{x^2}{2!}+o(x^2) \\ \arcsin(x)=x+\frac{x^3}{6}+o(x^3) \\ \tan(x)=x+\frac{x^3}{3}+o(x^3) $$

Answer 1

You received the good answer from @José Carlos Santos.

If I may suggest, when you have to compose Taylor series, work one piece at the time.

For example $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\arcsin(1-\cos(x))=\frac{x^2}{2}-\frac{x^4}{24}+\frac{x^6}{45}+O\left(x^8\right)$$ $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+\frac{17 x^7}{315}+O\left(x^9\right)$$ $$\tan^2(x)=x^2+\frac{2 x^4}{3}+\frac{17 x^6}{45}+\frac{62 x^8}{315}+O\left(x^{10}\right)$$ $$2\arcsin(1-\cos(x))-\tan^2(x)=-\frac{3 x^4}{4}-\frac{x^6}{3}+O\left(x^8\right)$$ $$\log(1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+O\left(x^{10}\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}-\frac{x^8}{315}+O\left(x^{10}\right)$$ $$\log(1+x^2)-\sin^2(x)=-\frac{x^4}{6}+\frac{13 x^6}{45}-\frac{311 x^8}{1260}+O\left(x^9\right)$$ $$\frac{2\arcsin(1-\cos(x))-\tan^2x}{\ln(1+x^2)-\sin^2(x)}=\frac{-\frac{3 x^4}{4}-\frac{x^6}{3}+O\left(x^8\right)} {-\frac{x^4}{6}+\frac{13 x^6}{45}-\frac{311 x^8}{1260}+O\left(x^9\right) }$$ Divide top an bottom by $x ^4$ and continue with long division to get $$\frac{2\arcsin(1-\cos(x))-\tan^2x}{\ln(1+x^2)-\sin^2(x)}=\frac{9}{2}+\frac{49 }{5}x^2+O\left(x^4\right)$$ which, for sure, shows the limit but moreover how it is approached.

Answer 2

Not quite. Since the Taylor series of the numerator begins with$$-\frac{3 x^4}{4}-\frac{x^6}{3}+\cdots$$and the Taylor series of the denominator begins with$$-\frac{x^4}{6}+\frac{13 x^6}{45}+\cdots,$$you should divide the numerator and the denominator by $x^4$, not $x^6$. You will get that the limit is$$\frac{-3/4}{-1/6}=\frac92.$$