# Find the limit

1. Aug 18, 2009

### yanjt

Hi,
I am wondering whether i answer the question correctly.
The question is,either find the limit of explain why there is no limit exists as k goes to infinity for
k sin(k$$\pi$$/3).

My answer for this question is that there is no limit exists since infinity multiply by any number will become infinity.Hence,regardless what value we get from sin(k$$\pi$$/3),it will become infinity when it is multiplied by infinity.Can I explain in this way?Thanks!

2. Aug 18, 2009

### JG89

Assuming by number you mean a constant value, then that is a good way to think about it intuitively. You know that for some values of k, sin(kpi/3) will be positive, and for other values of k it will be negative. k is always positive, so the expression k*sin(kpi/3) will switch between large positive values and large negative values, depending on the value of k.

If you need to make this into a rigorous argument, it shouldn't be too hard.

3. Aug 18, 2009

### yanjt

So there is no limit for this question?I can just write the statement out without showing any working?

4. Aug 18, 2009

### Дьявол

sin(kп/3) $\in$ [-1,1]

So k multiplied by any number in the interval [-1,1] (if k goes to infinity) is again infinity.

5. Aug 18, 2009

### Cyosis

This is simply not true, $-1*\infty=-\infty$. Also the OPs original argument is not true, because a limit resulting in infinity is defined just fine. The problem here as explained by JG80 is that the function oscillates and this oscillation goes on forever.

They ask for an explanation so yes explaining in words as to why there is no limit would be just fine.

6. Aug 18, 2009

### lanedance

one way you could look a litttle more rigourously is to pick some arbitrary montonic sequences {kn}, that tend to infinity. Then consider the sequences {f(kn)},

If you can find sequences with different limits, or a sequence that does not converge, you have shown the limit does not exist.

Or in similar fashion, u could find positivek,k' with N<k<k', such that f(k)>N, f(k')<-N, then as N is arbitrary, you have shown f(k) is unbounded from both above & below as k tends to infinity and hence has no limit, which is effectivey what cyosis & JG89 have said

Last edited: Aug 18, 2009
7. Aug 18, 2009

### Дьявол

I do not know what is wrong with my statement.
$-1*\infty$ equals infinity just negative. It is again infinity.

My statement was meant to express $x\in [-1,1] * \infty = \pm \infty$

By saying infinity, I mean $\pm \infty$.

Regards.

8. Aug 18, 2009

### Elucidus

There are errors in saying "infinity times a number is some infinity." The problem comes from the whether the other factor can equal zero or not or changes sign.

In the limit $$\lim_{x \rightarrow \infty}xe^{-x}$$ one could argue that since x approaches infinity and that e-x is a positve number then this limit diverges to infinity. Unfortunately this argument would be grossly wrong. The limit is in fact 0.

The issue with the limit being discussed is that the sine function is oscillating between -1 and 1 (and sometimes equalling 0). The explanation as to why the limit is undefined is that there is no finite number to which it converges and it does not diverge to either infinity. Both of these can be demonstrated by examing sequence of values that behave differently.

One needs to show why it does not converge and why it does not diverge to some infinity.

(BTW: a limit that diverges to infinity is still defined.)

--Elucidus

9. Aug 18, 2009

### HallsofIvy

The difficulty with saying $\pm \infty= \infty$ is that it mixes concepts of "infinity". The "one point compactification" of the real numbers adds the single point "$\infty$" and is topologically equivalent to a circle while the "Stone-Cech" compactification add $+\infty$ and $-\infty$ and is topologically equivalent to a closed line segment.

Also, some texts would say that {n} "diverges to infinity" while {-n} "diverges to negative infinity" and {$(-1)^{n}n$} simply "diverges".