1. I got this Limit problem i cannot solve, i would really appreciate if someone can help me to figure it out, thanks!

Lim (x/1-cos^2(x/2))
x-> 0

Thanks again!  2.

3. Since you cannot solve it i take that to mean that you have tried at least once? what approach did you use?  4. Yes indeed, i´ve tried not just once but many times, i tried using substitutions, identities, but non of them seems to work out.  5. Originally Posted by Cideos I got this Limit problem i cannot solve, i would really appreciate if someone can help me to figure it out, thanks!

Lim (x/1-cos^2(x/2))
x-> 0

Thanks again!
Comment 1: Should the expression be x/(1-cos^2(x/2)) ? As it stands it looks like x - cos^2(x/2).

Assuming that I have corrected it properly, have you used Hospital's rule?
Also you could simplify by 1 - cos^2(x/2) = sin^2(x/2).  6. Originally Posted by mathman  Originally Posted by Cideos I got this Limit problem i cannot solve, i would really appreciate if someone can help me to figure it out, thanks!

Lim (x/1-cos^2(x/2))
x-> 0

Thanks again!
Comment 1: Should the expression be x/(1-cos^2(x/2)) ? As it stands it looks like x - cos^2(x/2).

Assuming that I have corrected it properly, have you used Hospital's rule?
Also you could simplify by 1 - cos^2(x/2) = sin^2(x/2).
Yes, i didnt notice that mistake, and it is x/(1-cos^2(x/2)), thank you!.
I know i can simplify it with using many identities, but i have tried many times and in many ways and still cant solve it.

Also, one of the conditions my teacher gave me was to not use L'Hôpital's rule. I really would appreciate if you can help me!  7. The x part seems trivial, so we're looking at the 1-cos²(x/2).

Does 1-cos² remind you of anything?

And the x/2 argument makes me think of half-angle formulas.

Not that these two ideas must be resolved in the same step.

Hopefully it should start coming together in your mind.  8. Originally Posted by jrmonroe The x part seems trivial, so we're looking at the 1-cos²(x/2).

Does 1-cos² remind you of anything?

And the x/2 argument makes me think of half-angle formulas.

Not that these two ideas must be resolved in the same step.

Hopefully it should start coming together in your mind.
Thanks for the suggestion, but i´ve tried, as i wrote, with pitagoric identities, sum of angles, etc, and has not been of help.
But thanks for the interest.  9. Okay, then do the worst thing possible ... throw it into a spreadsheet, and see how it behaves. It approaches infinity, of course, but it does so in a particular way. This gives the answer, but is probably not the "work" your teacher wants to see.  Tags for this Thread
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