Showing that a limit exists

Printable View

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 $-1<x< 1$ , then $-1<x^2<1$
(3) As $n$ goes to infinity, $x^n$ goes to 0 if $-1<x<1$

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 :)

All times are GMT -8. The time now is 05:07 PM.