# Finding $\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}$ 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?

• using $\frac{\sin(x^2)}{\sin^2x}=\frac{\sin(x^2)}{x^2}\frac{x^2}{\sin^2x}$ help? – cand Jul 30 '14 at 15:51
$$\lim_{x\to0} \frac{\sin(x^2)}{\sin^2(x)}=\lim_{x\to0}\left(\frac{x}{\sin x}\right)^2 \frac{\sin(x^2)}{x^2}=1$$
$$\sin{x^2}=x^2+o(x^3)\\ \sin ^2 x=(x+o(x^2))^2=x^2+o(x^3) \implies\\ \frac{\sin x^2}{\sin ^2 x}=\frac{1+o(x)}{1+o(x)}\rightarrow 1 \text{ as } x\rightarrow 0$$