What is the limit of cosx as x goes to infinity?

Sep 26, 2015

There is no limit.

Explanation:

Recall or Note:

${\lim}_{x \rightarrow \infty} f \left(x\right) = L$ if and only if

for every positived $\epsilon$, there is an $M$ that satisfies:
for all $x > M$, $\left\mid f \left(x\right) - L \right\mid < \epsilon$

As $x$ increases without bound, $\cos x$ continues to attain every value between $- 1$ and $1$. So it cannot be getting and staying within $\epsilon$ of some one number, $L$,

Refer to explanation

Explanation:

Choose two sequences such as

${a}_{n} = 2 n \cdot \pi$ and ${b}_{n} = \left(2 n + 1\right) \cdot \pi$ where

${\lim}_{n \to \infty} {a}_{n} = \infty$ and ${\lim}_{n \to \infty} {b}_{n} = \infty$

But ${\lim}_{n \to \infty} \cos \left({a}_{n}\right) = \cos \left(2 \pi n\right) = 1$ and ${\lim}_{n \to \infty} \cos \left({b}_{n}\right) = \cos \left(\left(2 n + 1\right) \pi\right) = - 1$

Hence there is no limit for $\cos x$ as $x \to \infty$

The theorem that was used for the proof is

**(Divergence Criterion for Functional Limits):
Let f:A→R f:A→R, and let $c$ be a limit point of $A$. If there exist two sequences $\left({x}_{n}\right)$ and $\left({y}_{n}\right)$ in $A$ with x_n≠c and y_n≠c, and:
${\lim}_{n \to \infty} {x}_{n} = {\lim}_{n \to \infty} {y}_{n} = c$ but lim_(n->oo)f(x_n)≠lim_(n->oo)f(y_n)
then we can conclude that the functional lim_(x→c)f(x) does not exist.