Can the Limit of Sin(x) as x Approaches Infinity be Proven Graphically?

[pivoxa15]

Homework Statement

lim(x->infinity) sin(x)

The Attempt at a Solution

undefined.

 

[CompuChip]

Why do you think that?
Please give an argument.
And please say what knowledge you have of limits (basis calculus, real analysis, etc)

 

[ZioX]

Can you give a proof why this limit doesn't exist?

The sequential criterion for limits is handy, if you know this criterion.

The sequential criterion for limits is:

$\lim_{x \to a}f(x)=b$ iff for every sequence x_n in the extended real numbers (abstractly, the domain of f) converging to a has the property that f(x_n) converges to b.

Can you think of a sequence converging to infinity such that sin(x) converges? Can you think of another sequence converging to infinity such that sin(x) converges to a different value?

 

[Sartre]

I don't think you should just say that the limit is something or other. If you want to take limits in simple non-series functions, then you should look at the outer rims of the function.

For example sin(x) is defined as -1 < x < 1. Now you must work out if the limit is negative or positive infinity. And you should look at the definition that ZioX posted. It comes in handy.

 

[CompuChip]

Now you must work out if the limit is negative or positive infinity.

Or doesn't exist
Which is what he said (just didn't prove yet).

 

[HallsofIvy]

In other words, everyone is saying "yes, the limit doesn't exist, but you must tell why it doesn't exist"!

 

[Sartre]

Or doesn't exist
Which is what he said (just didn't prove yet).

Actually I reviewed this part of calculus a couple of days ago. So it is my bad om that one ;)

And the simplest proof of seeing that the limit doesn't exist is graphical.

 

[uiulic]

Ziox showed it doesn't exist.

x=2n*(pi)
x=(2n+1/2)*(pi)
when n->infinite, then got two different values.

 

[HallsofIvy]

For example sin(x) is defined as -1 < x < 1. Now you must work out if the limit is negative or positive infinity. And you should look at the definition that ZioX posted. It comes in handy.

This is non-sense. x can be any real number. I think what you meant was that if y= sin(x) then $-1\le y\le 1$. And you don't need to "work out if the limit is negative or positive infinity"- it's neither one, it just doesnt' exist.

 

[Gib Z]

The point is that the limit as x approaches infinity of finite period functions does not exist, as it will oscillate between the range of the function.

 

[CompuChip]

And the simplest proof of seeing that the limit doesn't exist is graphical.

Depends what you call a proof. And how rigorous a proof you want.