calculus - Finding $lim_{xto0} frac{sin(x^2)}{sin^2(x)}$ with Taylor series

Evaluate

$$\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}.$$

Using L'Hospital twice, I found this limit to be $1$. However, since the Taylor series expansions of $\sin(x^2)$ and $\sin^2(x)$ tell us that both of these approach $0$ like $x^2$, I'm wondering if we can argue that the limit must be 1 via the Taylor series formal way?

Answer

$$\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}=\lim_{x\to0}\left(\frac{x}{\sin x}\right)^2 \frac{\sin(x^2)}{x^2}=1$$