Thread: Disproving the existance of a limit at infinity

1. 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

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

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

Tonio

3. I'm afraid I haven't learned anything about sequences yet...is there another way?

4. 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 $x\to -\infty$ it is exactly the same for $x\to \infty$.
After all $(-\infty,0]~\&~[0,\infty)$ are 'copies' of one another as for as the sine function is concerned.

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

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

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

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

Tonio

6. 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?