2
$\begingroup$

How to evaluate this limit?

$$\lim_{x\to0}\frac{\sin x^2}{x^2} \sin \frac{1}{x}$$

L'Hospital not working

$\endgroup$
  • 8
    $\begingroup$ What is $\lim\limits_{z\rightarrow0}{\sin z\over z}$? Does $\lim\limits_{x\rightarrow0}\sin(1/x)$ exist? What can you conclude after answering these questions? $\endgroup$ – David Mitra Dec 10 '14 at 13:33
  • $\begingroup$ the limit does not exist since second one does not exist right @DavidMitra $\endgroup$ – Learnmore Dec 10 '14 at 13:36
  • $\begingroup$ @learningmaths Absolutely, Now you can post your own answer! Cheers! $\endgroup$ – Aditya Hase Dec 10 '14 at 13:39
  • $\begingroup$ Yes, indeed (and since the first limit I mentioned above does exist and is non-zero). $\endgroup$ – David Mitra Dec 10 '14 at 13:39
2
$\begingroup$

Hint:

$$\frac{\sin x^2}{x^2}$$ has a nonzero limit at $0$ (pretty easy to calculate). What is the behavior of $\sin\frac{1}{x}$ like when $x$ is small?

$\endgroup$

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.