- #1
yangxu
- 20
- 0
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.
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: