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

How to evaluate this limit?

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

L'Hospital not working

• 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? – David Mitra Dec 10 '14 at 13:33
• the limit does not exist since second one does not exist right @DavidMitra – Learnmore Dec 10 '14 at 13:36
• @learningmaths Absolutely, Now you can post your own answer! Cheers! – Aditya Hase Dec 10 '14 at 13:39
• Yes, indeed (and since the first limit I mentioned above does exist and is non-zero). – David Mitra Dec 10 '14 at 13:39

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