# Showing that a limit exists

• January 16th 2011, 03:47 AM
FGT12
Showing that a limit exists
How would you show that for every $x \in \mathbb{R}$, the limit

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

exists?
• January 16th 2011, 03:51 AM
Plato
Quote:

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

What is the question here?
• January 16th 2011, 03:52 AM
FGT12
Ive changed it
• January 16th 2011, 04:05 AM
DrSteve
Use the following facts:

(1) $-1\leq \cos x\leq 1$ for all $x$
(2) If $-1, then $-1
(3) As $n$ goes to infinity, $x^n$ goes to 0 if $-1

Now see if you can put the argument together.
• January 16th 2011, 04:07 AM
Prove It
Actually if $\displaystyle - 1 \leq x \leq 1$ then $\displaystyle 0 \leq x^2 \leq 1$.
• January 16th 2011, 05:07 AM
DrSteve
Thanks Prove It.

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