Okay, just so we're all operating on the same page, I'm assuming the problem statement said something like this: "Find the value(s) of n, such that the following limit exists."
\(\displaystyle \displaystyle \lim _{x\to 0}\left(\frac{1-cos\left(2x\right)}{x^n}\right)\)
If the above is an incorrect summary of the given problem, please reply with the actual problem, quoting word-for-word if possible. Also, please show all of your work on this problem, even if you know it's wrong. If you're stuck at the beginning and have no work to show, here's a few hints to maybe get you thinking in the right direction:
Do you, perhaps, know a trig identity to rewrite 1 - cos(2x) as a single term? How does applying this identity affect the limit?
This limit may prove helpful: \(\displaystyle \displaystyle \lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=1\)
Recall that you can split apart a limit involving multiplication into multiple limits. i.e. \(\displaystyle \displaystyle \lim _{x\to 0}\left(a\cdot b\cdot c\right)=\lim _{x\to 0}\left(a\right)\cdot \lim _{x\to 0}\left(b\right)\cdot \lim _{x\to 0}\left(c\right)\)
