3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Short Answer: yes. $\endgroup$ – Brad Jul 30 '14 at 15:50
  • 1
    $\begingroup$ using $\frac{\sin(x^2)}{\sin^2x}=\frac{\sin(x^2)}{x^2}\frac{x^2}{\sin^2x}$ help? $\endgroup$ – cand Jul 30 '14 at 15:51
3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ (+1)much better than Taylor expansion! $\endgroup$ – Alex Jul 30 '14 at 16:25
3
$\begingroup$

$$\sin{x^2}=x^2+o(x^3)\\ \sin ^2 x=(x+o(x^2))^2=x^2+o(x^3) \implies\\ \frac{\sin x^2}{\sin ^2 x}=\frac{1+o(x)}{1+o(x)}\rightarrow 1 \text{ as } x\rightarrow 0$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.