# Thread: Does the limit as x-> inifinity of cosx exist?

1. ## Does the limit as x-> inifinity of cosx exist?

lim x-> infinity (e^(-x)cosx) ?

First of all does the limit as x-> inifinity of cosx even exist?
I thought it would = inifinity but then again i feel like it DNE as well.

Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists

2. Originally Posted by kensington
lim x-> infinity (e^(-x)cosx) ?

First of all does the limit as x-> inifinity of cosx even exist?
I thought it would = inifinity but then again i feel like it DNE as well.

Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists

$\displaystyle \displaystyle \lim_{x \to \infty} e^{-x} \cos{x} = 0$

the fact that $\displaystyle \displaystyle \lim_{x \to \infty} \cos{x}$ does not exist is inconsequential in this case ... the value of $\displaystyle \cos{x}$ stays relatively fixed between -1 and 1 ... $\displaystyle e^{-x}$ is the dominant factor here.

3. Originally Posted by kensington
lim x-> infinity (e^(-x)cosx) ?

First of all does the limit as x-> inifinity of cosx even exist?
I thought it would = inifinity but then again i feel like it DNE as well.

Ive atempted this problem in 2 ways, but i first would like to see if cosx as x approaches infinity even exists
The trick here is to recall that $\displaystyle \forall x\in\mathbb{R},\,-1\leq \cos x\leq 1$. This is important because as $\displaystyle x\to\infty$, $\displaystyle \cos x$ is going to remain finite (but we don't necessarily know the limit value).

So the only term that will really effect the limit will be the $\displaystyle e^{-x}$ term. So we should see that $\displaystyle \lim\limits_{x\to\infty}e^{-x}\cos x\rightarrow 0\cdot (\text{finite number}) = 0$.

Does this make sense?

EDIT: skeeter beat me to it.

4. Originally Posted by Chris L T521
The trick here is to recall that $\displaystyle \forall x\in\mathbb{R},\,-1\leq \cos x\leq 1$. This is important because as $\displaystyle x\to\infty$, $\displaystyle \cos x$ is going to remain finite (but we don't necessarily know the limit value).

So the only term that will really effect the limit will be the $\displaystyle e^{-x}$ term. So we should see that $\displaystyle \lim\limits_{x\to\infty}e^{-x}\cos x\rightarrow 0\cdot (\text{finite number}) = 0$.

Does this make sense?
Ohh because i was simply trying to figure out if we use L'hospitals rule? or you can even rewrite is as cosx/e^x but i was always stuck with the cosx and then e^x would approach infinity in that case.
So you mierly just "ignored" the cos and did the limit to e^-x which is 0?

5. Originally Posted by kensington
Ohh because i was simply trying to figure out if we use L'hospitals rule? or you can even rewrite is as cosx/e^x but i was always stuck with the cosx and then e^x would approach infinity in that case.
So you mierly just "ignored" the cos and did the limit to e^-x which is 0?
I wouldn't call it "ignore" ... it would be easier to see the idea if you wrote the limit as

$\displaystyle \displaystyle \lim_{x \to \infty} \frac{\cos{x}}{e^x}$

as $\displaystyle x \to \infty$ , $\displaystyle \cos{x}$ oscillates between -1 and 1 (i.e. stays relatively fixed) and $\displaystyle e^x$ gets large w/o bound ...

$\displaystyle \displaystyle \frac{value \, between \, -1 \, and \, 1}{very \, large \, value} \to 0$

6. Originally Posted by skeeter
I wouldn't call it "ignore" ... it would be easier to see the idea if you wrote the limit as

$\displaystyle \displaystyle \lim_{x \to \infty} \frac{\cos{x}}{e^x}$

as $\displaystyle x \to \infty$ , $\displaystyle \cos{x}$ oscillates between -1 and 1 (i.e. stays relatively fixed) and $\displaystyle e^x$ gets large w/o bound ...

$\displaystyle \displaystyle \frac{value \, between \, -1 \, and \, 1}{very \, large \, value} \to 0$

Got it thanks!