<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] \begin{align*}
\lim_{x\rightarrow 0}\dfrac{x^{2}-\log(1+x^{2})}{x^{2}\sin^{2}x}&=\lim_{x\rightarrow 0}\dfrac{x^{2}-x^{2}+\dfrac{1}{2}x^{4}-\dfrac{1}{3}x^{6}\cdots}{x^{2}\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}-\dfrac{1}{3}x^{4}\cdots}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\left(\dfrac{1}{2}\cdot\dfrac{x^{2}}{\sin^{2}x}\right)\cdot\lim_{x\rightarrow 0}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)\\
&=\dfrac{1}{2}.
\end{align*}
</s>