How would you show that for every $\displaystyle x \in \mathbb{R}$, the limit

$\displaystyle \lim_{m\to \infty} (cos(x))^{2m}$

exists?

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- Jan 16th 2011, 03:47 AMFGT12Showing that a limit exists
How would you show that for every $\displaystyle x \in \mathbb{R}$, the limit

$\displaystyle \lim_{m\to \infty} (cos(x))^{2m}$

exists? - Jan 16th 2011, 03:51 AMPlato
- Jan 16th 2011, 03:52 AMFGT12
Ive changed it

- Jan 16th 2011, 04:05 AMDrSteve
Use the following facts:

(1) $\displaystyle -1\leq \cos x\leq 1$ for all $\displaystyle x$

(2) If $\displaystyle -1<x< 1$, then $\displaystyle -1<x^2<1$

(3) As $\displaystyle n$ goes to infinity, $\displaystyle x^n$ goes to 0 if $\displaystyle -1<x<1$

Now see if you can put the argument together. - Jan 16th 2011, 04:07 AMProve It
Actually if $\displaystyle \displaystyle - 1 \leq x \leq 1$ then $\displaystyle \displaystyle 0 \leq x^2 \leq 1$.

- Jan 16th 2011, 05:07 AMDrSteve
Thanks Prove It.

Although technically everything I've written is true and is sufficient to finish the argument :)