# Having trouble finding this limit

## Main Question or Discussion Point

Hi all,

My girlfriend asked me this question just now, however I have no idea how I should approach to solve it, I highly appreciate if anyone could shed lights on this:

lim x^2 times cos^2(x^-2)
x->0

I tried using the squeeze theorem:

-1 < cos(1/x) < 1
thus:
-x^2 < x^2 (cos(1/x)) < x^2
or
-x^2 cos(1/x) < x^2 (cos^2(x^-2)) < x^2 cos(1/x)
Therefore, as x->0 in the middle, the two sides also approach 0.

But I don't think it makes any sense... since 1/x as x-> 0 cannot really be used as part of the intervals representing -1 and 1.

I also tried rearranging cos^2 (x^-2), but I don't think it's any use.

Last edited:

calc 1 or 2?

$$x^2\cos^2({\frac{1}{x^2}})$$

-1<cos(theta) <1 for any value of theta

so x^2 * cos (theta)->0 as x -> 0

What's wrong with that?

HallsofIvy
Homework Helper
$-1\le cos(\theta)\le 1$ even when $\theta= 1/x^2$! That's christianjb's point.

Thank you for your replies, guys.

rocophysics:
Yes, that's the correct output of the question. I haven't done calculus for a long time, but I believe it's calc 1.

christianjib & HallsofIvy:
But if it's cos^2 (x^-2), can you still apply the same rule?

HallsofIvy
Actually, in that case, it's easier: $0\le cos^2(\theta)\le 1$ for all $\theta$!
Actually, in that case, it's easier: $0\le cos^2(\theta)\le 1$ for all $\theta$!