What is the Limit of Xcos(1/x) at 0?

[kathrynag]

Homework Statement

Define f: (0,1)---->R by . Does f have a limit at 0?

Homework Equations

0<absvalue(x-x_0)<delta
absvalue(f(x)-L)<epsilon

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)?

 

[Mark44]

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?

 

[kathrynag]

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?

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

 

[HallsofIvy]

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

 

[D H]

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.

 

[Mark44]

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.

 

[kathrynag]

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.

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...

 

[kathrynag]

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

 

[HallsofIvy]

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...

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?

 

[kathrynag]

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?

Let delta =epsilon?

 

[kathrynag]

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

 

[Mark44]

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

 

[kathrynag]

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

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

 

[HallsofIvy]

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

 

[kathrynag]

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

 

[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.

 

[kathrynag]

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

 

[kathrynag]

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