# Non-existing limits

1. Oct 2, 2004

### Dr-NiKoN

How do you show that a limit doesn't exist?

$\lim_{x\to 0}cos(\frac{1}{x})$

This doesn't have any limit, since it will go from -1 to 1 when x goes towards 0, but how do I show that mathematically?

2. Oct 2, 2004

### arildno

Choose an arbitrary number called "L". (Think of it as between -1 and 1 if you like)
Show that there exists a neighbourhood U of 0, so that given |L|<1, there will always be some x in U so that |L-cos(1/x)|>1/2

3. Oct 2, 2004

### modmans2ndcoming

or, if say he/she has not had analytical training yet, he/she could just set that equal to the empty set since the solution set is empty.

4. Oct 3, 2004

### cogito²

find two sequences that approach 0 and then show that the functions of those sequences do not approach the same number. an easy way would be to find a sequence that goes to y = 1 and another to y = 0. if you don't understand then just say so.

5. Oct 3, 2004

### Dr-NiKoN

Is that valid proof?

Couldn't a function have such a curve that at one point it goes towards 0 and a later point goes towards 1 and then at a later point goes towards 0 and continues that way as x goes towards inf?

Maybe I misunderstood. We are not supposed to show this yet, so It's probably something we will learn trough analysis later.

6. Oct 3, 2004

### arildno

This is certainly a correct proof:
Every sequence converging to x0 will have members arbitrarily close to x0.
Let Yn, Xn be two such sequences.
Evidently, if:
$$\lim_{n\to\infty}(f(X_{n})-f(Y_{n}))\neq0$$
then f cannot have a limit at x0

7. Oct 3, 2004

### shmoe

I think you misunderstood slightly, it's important that both of the sequences are converging to 0, but the function's values on these sequences converge to different values.

More precisely, we want two sequences, $$x_1,x_2,\ldots$$ and $$y_1,y_2,\ldots$$ where $$\lim_{n\to 0}x_n=\lim_{n\to 0}y_n=0$$ but $$\lim_{n\to 0}f(x_n)=1\neq 0=\lim_{n\to 0}f(y_n)$$

(1 and 0 were chosen here because they are convenient for this specific problem)

This is certainly a correct way of doing things. If a function has a limit at zero, than on any sequence converging to zero, the values of the function along that sequence must equal this limit (this is worth trying to prove if you can).

8. Oct 3, 2004

### arildno

I would like to add that cogito's and shmoe's way is often the simplest manner to complete a proof.

9. Oct 3, 2004

### Dr-NiKoN

Ok, here is a real-life problem I'm dealing with.
$f(x) = \frac{1}{x}\sin(\frac{1}{x})$
How many times will f(x) = 0 in the range [0, ->].

I have:
(Check if it touches 0 as x goes towards inf.)
$f(x) = \lim_{x\to \infty}\frac{1}{x}$
$g(x) = \lim_{x\to \infty}\sin(\frac{1}{x})$

$\lim_{x\to \infty}f(x) + g(x) = 0$

Thus, no touching that way.
(Check if x touches 0 as x goes towards 0)
$f(x) = \lim_{x\to 0}\frac{1}{x}$
$g(x) = \lim_{x\to 0}\sin(\frac{1}{x})$

$\lim_{x\to 0}f(x) + g(x) = undefined$

Is this good enough proof that:
$f(x) = \frac{1}{x}\sin(\frac{1}{x})$
Will touch 0 an inifite amount of times in the range [0, ->]?
Or is it finite, because x = 0 at some point?

Last edited: Oct 3, 2004
10. Oct 3, 2004

### arildno

It's a lot simpler:
Note that sin(y)=0, then $$y=n\pi$$ for some integer n.

Hence, at $$x=\frac{1}{n\pi}$$, you'll have the zeroes f
(an infinite amount of them)

11. Oct 3, 2004

### Dr-NiKoN

Damn, there's so many different variations in calculus. Seems you need to keep an eye up for almost anything.

Thanks though, it makes perfect sense, I'm just bitter I didn't see it myself :)

12. Oct 3, 2004

### arildno

And there's the calculus of variations, as well..