# Xcos(1/x) limit at 0

1. Nov 13, 2008

### kathrynag

1. The problem statement, all variables and given/known data
Define f: (0,1)---->R by . Does f have a limit at 0?

2. Relevant equations
0<absvalue(x-x_0)<delta
absvalue(f(x)-L)<epsilon

3. The attempt at a solution
epsilon is arbitrary.
0<absvalue(x-0)<delta. Then absvalue(xcos1/x-0)<epsilon.
0<absvalue(x)<delta
absvalue(xcos1/x)<epsilon
absvalue(x)absvalue(cos1/x)<epsilon

This is where I get stuck? Do I divide by absvalue(cos1/x) and let delta= sec(1/x)?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 13, 2008

### Staff: Mentor

I think you are confused between the ideas of finding the limit of a function and proving (with epsilons and deltas) that a certain number is the limit of that function.

The question is "Does f have a limit at 0?" The answer should be either "yes" or "no." From your work it would appear that you believe the answer is yes. You can justify this belief with this inequality:
$$0 \leq x cos(\frac{1}{x}) \leq x$$

As x approaches 0, what happens to each end of the inequality above?

3. Nov 13, 2008

### kathrynag

No, I am supposed to prove it. I'm just not sure about my proof.

4. Nov 13, 2008

### HallsofIvy

Staff Emeritus
A proof does not necessarily mean an "epsilon- delta" proof. If you can use the "sandwiching" theorem, that would work nicely.

5. Nov 13, 2008

### D H

Staff Emeritus
Replace that nasty cos(1/x) term with something much simpler. If you can find some f(x) such that |cos(1/x)| < |f(x)| you can use this function in lieu of cos(1/x).

Note: You should prove this or find the theorem in your text that proves this for you.

6. Nov 13, 2008

### Staff: Mentor

You didn't include that in your problem statement.

Little tip: instead of typing absvalue over and over, use the | key. It'll save you a lot of typing.

So, given an epsilon, you need to find a delta so that x < delta ==> |x cos(1/x)| < epsilon. Something that might be helpful is that |x cos(1/x)| <= |x| for all x != 0. Maybe you can use this idea.

7. Nov 13, 2008

### kathrynag

So, I have lxcos(1/x)-0l < epsilon
lxcos(1/x)l < epsilon
Then lxl lcos(1/x)l < epsilon
But lxcos(1/x)l < lxl

I'm not sure where to go with this idea....

8. Nov 13, 2008

### kathrynag

So, could I then say since lxl<epsilom, that delta=1/lxl?

9. Nov 13, 2008

### HallsofIvy

Staff Emeritus
Well, that says that if |x|< epsilon, then |x cos(1/x)|, which is even smaller, will be less than epsilon. How do you choose delta to be sure |x|< epsilon?

10. Nov 13, 2008

### kathrynag

Let delta =epsilon?

11. Nov 13, 2008

### kathrynag

Let delta=epsilon because if epsilon =delta then lxcos(1/x)l < delta?

12. Nov 13, 2008

### Staff: Mentor

Yes, that's where Halls and I were heading.

13. Nov 13, 2008

### kathrynag

Ok, so it's plain delta = epsilon? Just that simple? I wasn't sure since I did another problem where I had delta=epsilon/2

14. Nov 14, 2008

### HallsofIvy

Staff Emeritus
You can't expect all problems to have the same answer!

15. Nov 14, 2008

### kathrynag

Ok, thanks. i just thought that seemed a little too easy, I guess, haha.

16. Nov 14, 2008

### snipez90

If you have to prove a limit which involves trig functions, an epsilon-delta argument usually works nicely. The absolute values give a nice bound on sin and cos functions. The squeeze theorem is useful as well, but you should try deriving it on your own using the epsilon-delta.

17. Nov 19, 2008

### kathrynag

Ok, I'm unsure on the step where I say $$\left|x\right|<\epsilon$$. How do I get this?

18. Nov 20, 2008

### kathrynag

Anybody have any hints on where this comes from? like I know lxl < delta.