# Disproving the existance of a limit at infinity

• Mar 26th 2011, 06:16 AM
moses
Disproving the existance of a limit at infinity
Hello

I'm given the function x*sin(x), and asked whether it has a limit of some sort as x goes to negative infinity. In other words does it converge to a number or go off to positive infinity or negative infinity.

Based on the graph of the function my intuition is that it doesn't have a limit of any sort since it is constantly fluctuating between very large and very small values.

I think i've managed to prove that it doesn't converge to any number at negative infinity, because on any interval can always supply an x1 = -pi/2 -n*pi for which the value is -x1 and an x2 = -(3/2) * pi -n*pi for which the value is x2, i.e. the function doesn't get closer to any single number.

But I'm having a lot of trouble with the infinite limits, proving that the function doesn't go off to positive or negative infinity...help would be much appreciated :)
• Mar 26th 2011, 09:01 AM
tonio
Quote:

Originally Posted by moses
Hello

I'm given the function x*sin(x), and asked whether it has a limit of some sort as x goes to negative infinity. In other words does it converge to a number or go off to positive infinity or negative infinity.

Based on the graph of the function my intuition is that it doesn't have a limit of any sort since it is constantly fluctuating between very large and very small values.

I think i've managed to prove that it doesn't converge to any number at negative infinity, because on any interval can always supply an x1 = -pi/2 -n*pi for which the value is -x1 and an x2 = -(3/2) * pi -n*pi for which the value is x2, i.e. the function doesn't get closer to any single number.

But I'm having a lot of trouble with the infinite limits, proving that the function doesn't go off to positive or negative infinity...help would be much appreciated :)

Take the sequences $\displaystyle \displaystyle{\{-n\pi\}\,,\,\left\{-(4n-1)\frac{\pi}{2}\right\}$ . We know that if $\displaystyle \lim\limits_{x\to -\infty}x\sin x$ exists, then it is

the same no matter how we choose to make $\displaystyle x\to -\infty$ , so do this through the above two seq's.

Tonio
• Mar 26th 2011, 10:39 AM
moses
I'm afraid I haven't learned anything about sequences yet...is there another way?
• Mar 26th 2011, 11:53 AM
Plato
Quote:

Originally Posted by moses
I'm afraid I haven't learned anything about sequences yet...is there another way?

I am truly puzzled by this question.
If you can do it for $\displaystyle x\to -\infty$ it is exactly the same for $\displaystyle x\to \infty$.
After all $\displaystyle (-\infty,0]~\&~[0,\infty)$ are 'copies' of one another as for as the sine function is concerned.
• Mar 26th 2011, 12:41 PM
tonio
Quote:

Originally Posted by Plato
I am truly puzzled by this question.
If you can do it for $\displaystyle x\to -\infty$ it is exactly the same for $\displaystyle x\to \infty$.
After all $\displaystyle (-\infty,0]~\&~[0,\infty)$ are 'copies' of one another as for as the sine function is concerned.

It probably is a witty question: prove that $\displaystyle \lim\limits_{x\to -\infty}x\sin x=\lim\limits_{x\to\infty}x\sin x$ , since the

function $\displaystyle x\sin x$ is even...

Anyway, I can't see right now a straightforward way to prove what the OP wants with resourcing to sequences.

Tonio
• Mar 26th 2011, 09:53 PM
moses
Okay, I see I phrased the question pretty unclearly

Basically what I meant was, how can I prove that the limit of xsinx as x goes to negative infinity is not positive infinity, and also that the limit of xsinx as x goes to negative infinity is not negative infinity?