# Find the limit of $\sin^2(1/x^2)$ when $x\to 0$ or show the limit doesn't exist

Find the limit or show the limit doesn't exist for

$$\lim_{x\to 0} \sin^2 \frac{1}{x^2}$$

I'm relatively new to limits and i'm unsure of how to show this.

I thought of breaking it down further to

$$\lim_{x\to 0} \sin^2 \frac{1}{x^2}=\bigl(\lim_{x\to 0} \sin \frac{1}{x^2} \bigr)^2$$ but I'm stuck here.

• Hint: Does the limit $\lim\limits_{n\to\infty}\sin n$ exist? (Recall that $\sin$ keeps oscillating between $-1$ and $1$) – Workaholic Apr 1 '16 at 10:06
• $n=x$? $0=\infty$? – Did Apr 1 '16 at 10:07
• Do you mean x instead of n – Archis Welankar Apr 1 '16 at 10:07
• @Workaholic More than that is necessary to conclude (otherwise, $\sin(\pi n)$ would diverge as well). – Did Apr 1 '16 at 10:07
• Yeah sorry it's $x$. have been working on so many problems my brain ain't thinking straight. thanks! – Danxe Apr 1 '16 at 10:08

## 1 Answer

$x=\dfrac2{\sqrt{\pi*k}}$ then $\sin^2(\dfrac1{x^2})=\sin^2(\dfrac{\pi*k}4)=0,\dfrac12,1,\dfrac12,0,\dfrac12$,...