Prove that the limit of sin(sqrt(x+1))-sin(sqrt(x-1)) at infinity doesn't exist

[AdrianZ]

Homework Statement

The problem is to prove that the limit of sin(sqrt(x+1)) - sin(sqrt(x-1)) when x goes to infinity doesn't exist.

The Attempt at a Solution

well, I converted sin(sqrt(x+1)) - sin(sqrt(x-1)) into the alternative form -2sin(sqrt(x+1)/2 - sqrt(x-1)/2)cos(sqrt(x+1)/2+sqrt(x-1)/2). Now my question is to show that limcos(x) as x goes to infinity doesn't exist. I tried to use epsilon-delta definition to prove it but I failed. I know that cos(x) is a periodic function and oscillates between 1 and -1 and I do know that its limit at infinity doesn't exist for that reason, but I want a mathematical proof that uses definitions or theorems to show that.
Thanks in advance.

 

[Hurkyl]

I tried to use epsilon-delta definition to prove it but I failed. I know that cos(x) is a periodic function and oscillates between 1 and -1

What did you try for the epsilon-delta definition? What was the intuitive idea behind it? (it's that cos(x) oscillates from 1 and -1) How did you try to translate the idea into epsilon-"delta" form?


Did you actually write down correctly what you need to prove?

For every L, there exists $\epsilon > 0$ such that for every $N > 0$ there exists an $x > N$ such that $|x - L| \geq \epsilon$​

(Do you understand why this is what you need to prove?)

 

[micromass]

Homework Statement

The problem is to prove that the limit of sin(sqrt(x+1)) - sin(sqrt(x-1)) when x goes to infinity doesn't exist.

The Attempt at a Solution

well, I converted sin(sqrt(x+1)) - sin(sqrt(x-1)) into the alternative form -2sin(sqrt(x+1)/2 - sqrt(x-1)/2)cos(sqrt(x+1)/2+sqrt(x-1)/2).

What is the limit of

$$\frac{\sqrt{x+1}-\sqrt{x-1}}{2}$$

as x goes to infinity?

 

[AdrianZ]

What did you try for the epsilon-delta definition? What was the intuitive idea behind it? (it's that cos(x) oscillates from 1 and -1) How did you try to translate the idea into epsilon-"delta" form?Did you actually write down correctly what you need to prove?

For every L, there exists $\epsilon > 0$ such that for every $N > 0$ there exists an $x > N$ such that $|x - L| \geq \epsilon$​

(Do you understand why this is what you need to prove?)

well, I assumed that the limit exists and is equal to L, I wrote down the definition that there exists a positive N that for any positive epsilon if x>N then |f(x) -L| < epsilon. then I tried to come to a contradiction. Actually I came to a contradiction using the fact that |cos(x)|<=1. but my argument didn't use any of my assumptions, so I guess It must be a wrong argument because if my argument were true, then the same argument could be used to show that the limit of cos(x) doesn't exist at any point which is obviously a wrong statement.
and Yes, I understand how you come up with that logical statement and why I need to prove that. That's simply the negation of the definition of a limit at infinity. I know what I need to prove, but I don't know what definitions or theorems I should start with.

What is the limit of

$$\frac{\sqrt{x+1}-\sqrt{x-1}}{2}$$

as x goes to infinity?

I guess it's zero. therefore $$sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})$$is zero. So?

 

[micromass]

I guess it's zero. therefore $$sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})$$is zero. So?

If that goes to zero, then what does

$$-2\sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})\cos(\frac{\sqrt{x+1}+\sqrt{x-1}}{2})$$

go to? Apply the squeeze theorem.

 

[AdrianZ]

hmmm. I don't know. the first limit goes to zero, but the cosine part does not exist. How can I apply the squeeze theorem to show that product of two limits doesn't exist? I thought I could apply it only to cases like limf(x)<=limg(x)<=limh(x) as x approaches to a where limf(x)=limh(x)=A and then I could conclude that limg(x) when x approaches to a goes to A.

I found a proof to show that the cosine function doesn't have a limit at infinity. here's my proof:
Assume that it has a limit at infinity that is equal to L. by definition, We must show that there exists N>0 that for any epsilon>0 we have: if x>N then |cos(x) - L|<epsilon.
Now let's choose epsilon to be 0.5 for example. take two different arbitrarily large values for x. let's take x to be in the set {2(pi)n or 2(pi)n+(pi/2): n is in positive integers}. as N goes to infinity, we can always find a member x in the set that is larger than N. so assume that for some n=k, we've chosen x to be equal to 2(pi)k and then equal to 2(pi)k+(pi/2). Now:
|cos(x) -L| = | cos(2pik) - L| < 0.5 => | 1 - L | < 0.5 => 0.5 < L < 1.5
|cos(x) -L| = | cos(2pik+pi/2) - L| < 0.5 => |0 - L| < 0.5 => -0.5<L<0.5
And these two inequalities don't include any L. therefore no such L exists.
Is this a convincing proof?

Now I have to show that the product of two limits at infinity,when one of them doesn't converge, does not exist. lol

 

[micromass]

Maybe you can do

$$-\sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})\leq\sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})\cos(\frac{\sqrt{x+1}+\sqrt{x-1}}{2})\leq \sin(\frac{\sqrt{x+1}-\sqrt{x-1}}{2})$$

And use the squeeze theorem on this.

 

[AdrianZ]

hmmm, I think you're right because whatever the limit of the cosine part is, It's something between -1 and +1 and if we multiply this by the limit of something which is zero it should go to zero, but I'm not sure if we are allowed to do that because the cosine part doesn't exist. but you made a very good point. Actually the question asks us to prove that the limit doesn't exist but You're saying that the limit exists and is equal to zero. right?

Anyway, Is my proof correct?

 

[micromass]

hmmm, I think you're right because whatever the limit of the cosine part is, It's something between -1 and +1 and if we multiply this by the limit of something which is zero it should go to zero, but I'm not sure if we are allowed to do that because the cosine part doesn't exist. but you made a very good point. Actually the question asks us to prove that the limit doesn't exist but You're saying that the limit exists and is equal to zero. right?

Indeed, I claim that the limit is 0. See also http://www.wolframalpha.com/input/?...mit.limitvariable--.**Limit.direction---.*--"

 

[micromass]

I found a proof to show that the cosine function doesn't have a limit at infinity. here's my proof:
Assume that it has a limit at infinity that is equal to L. by definition, We must show that there exists N>0 that for any epsilon>0 we have: if x>N then |cos(x) - L|<epsilon.
Now let's choose epsilon to be 0.5 for example. take two different arbitrarily large values for x. let's take x to be in the set {2(pi)n or 2(pi)n+(pi/2): n is in positive integers}. as N goes to infinity, we can always find a member x in the set that is larger than N. so assume that for some n=k, we've chosen x to be equal to 2(pi)k and then equal to 2(pi)k+(pi/2). Now:
|cos(x) -L| = | cos(2pik) - L| < 0.5 => | 1 - L | < 0.5 => 0.5 < L < 1.5
|cos(x) -L| = | cos(2pik+pi/2) - L| < 0.5 => |0 - L| < 0.5 => -0.5<L<0.5
And these two inequalities don't include any L. therefore no such L exists.
Is this a convincing proof?

This is a nice proof that cos(x) doesn't have a limit at infinity!

 

[AdrianZ]

Indeed, I claim that the limit is 0. See also http://www.wolframalpha.com/input/?...mit.limitvariable--.**Limit.direction---.*--"

You know, the first time that I tried to prove that the limit doesn't exist I used this trigonometric inequality that |sin(y)-sin(x)|<=|y-x| and then I concluded that the limit must approach zero at infinity but I thought that I had made a mistake somewhere in my proof. Now I understand that I had been wasting my time from two days ago to now to prove a false statement. lol. I should check to be sure if our math professor has passed his under-graduate courses before he receives his Phd. lol. Thank you for your help micromass.