Showing 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?

Quote:

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Originally Posted by FGT12

How would show that for ever $\displaystyle x \in \mathbb{R}$, the limit $\displaystyle \lim_{m\to \infty} (cos(x))^{2m}$

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What is the question here?

Ive changed it

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.

Actually if $\displaystyle \displaystyle - 1 \leq x \leq 1$ then $\displaystyle \displaystyle 0 \leq x^2 \leq 1$.

Thanks Prove It.

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