0
$\begingroup$

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.

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

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

$\endgroup$

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.