<s>[INST] Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...  
 [/INST] Note that by Taylor's series


*

*$\log(1+x^2)=x^2-\frac{x^4}{2}+o(x^4)$

*$\sin x^2=x^2+o(x^2)$


thus
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-x^2+\frac{x^4}{2}+o(x^4)}{x^4+o(x^4)}=\frac{\frac12+o(1)}{1+o(1)}\to \frac12$$
</s>