Having trouble finding this limit

  • Thread starter yangxu
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  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
1,752
1
calc 1 or 2?

is your problem

[tex]x^2\cos^2({\frac{1}{x^2}})[/tex]
 
  • #3
529
1
-1<cos(theta) <1 for any value of theta

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

What's wrong with that?
 
  • #4
HallsofIvy
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[itex]-1\le cos(\theta)\le 1[/itex] even when [itex]\theta= 1/x^2[/itex]! That's christianjb's point.
 
  • #5
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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?
 
  • #6
HallsofIvy
Science Advisor
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Actually, in that case, it's easier: [itex]0\le cos^2(\theta)\le 1[/itex] for all [itex]\theta[/itex]!
 
  • #7
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Actually, in that case, it's easier: [itex]0\le cos^2(\theta)\le 1[/itex] for all [itex]\theta[/itex]!
Wow, never thought of that, thanks so much HallsofIvy. :D
 

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