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...  

$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{\dfrac 1{x^2}-\dfrac{\log(1+x^2)}{x^4}}{\dfrac{\sin^2x}{x^2}}$$
Taking $z=x^2$, note that the numerator becomes $$\dfrac 1z-\dfrac{\log(1+z)}{z^2}=\dfrac{z-\log(1+z)}{z^2}\to\dfrac{1-\frac 1{1+z}}{2z}\to\dfrac{1}{2(z+1)^2}\to\frac 12$$
as $z=x^2\to 0$ as $x\to 0$ where the final arrowed steps use L'Hopital
The denominator goes to $1$ since $\dfrac{\sin x}x\to 1$ as $x\to 0$
Thus, the limit is $1/2$ by a final application of L'Hopital's rule.
