How Do You Solve This Tricky Limit Involving Cosine and x Squared?

[yangxu]

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.

Please enlighten on this, thanks in advance.

 

[rocomath]

calc 1 or 2?

is your problem

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

 

[christianjb]

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

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

What's wrong with that?

 

[HallsofIvy]

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

 

[yangxu]

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!

 

[yangxu]

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

Wow, never thought of that, thanks so much HallsofIvy. :D