# Showing that a limit exists

• Jan 16th 2011, 03:47 AM
FGT12
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?
• Jan 16th 2011, 03:51 AM
Plato
Quote:

Originally Posted by FGT12
How would show that for ever $\displaystyle x \in \mathbb{R}$, the limit $\displaystyle \lim_{m\to \infty} (cos(x))^{2m}$

What is the question here?
• Jan 16th 2011, 03:52 AM
FGT12
Ive changed it
• Jan 16th 2011, 04:05 AM
DrSteve
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 AM
Prove It
Actually if $\displaystyle \displaystyle - 1 \leq x \leq 1$ then $\displaystyle \displaystyle 0 \leq x^2 \leq 1$.
• Jan 16th 2011, 05:07 AM
DrSteve
Thanks Prove It.

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