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...  

As an alternative
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-\log(1+x^2)}{x^4}\cdot\frac{x^2}{\sin^2x}\to \frac12$$
indeed
$\frac{x^2}{\sin^2x}\to 1$ by standard limit
and let $y=x^2\to 0$
$$\frac{x^2-\log(1+x^2)}{x^4}=\frac{y-\log(1+y)}{y^2}\stackrel{HR}\implies\frac{1-\frac1{1+y}}{2y}=\frac{1+y-1}{2y(1+y)}=\frac{1}{2(1+y)}\to\frac12$$
