How to Prove the Squeeze Theorem for a Tricky Limit?

[realism877]

Prove that the lim approaches 0 from the right (sqr(x))(1+sin2((2pi)/(x))=0

The 2 indicates sin to the second power.

 

[Mark44]

Prove that the lim approaches 0 from the right (sqr(x))(1+sin2((2pi)/(x))=0

The 2 indicates sin to the second power.

Here's your limit, in full LaTeX glory:
\lim_{x \to 0^+}\sqrt{x}(1 + sin^2(2\pi/x))

If you're curious as to how I did it, click the expression.

The key here is realizing that 0 <= sin²(u) <= 1, for all real u.

 

[realism877]

How to I solve this from the right side only?

 

[Mark44]

The reason for the limit being a one-sided limit is that \sqrt{x} isn't defined for negative values of x.

What you need to do in this problem is to write an inequality like this:
A \leq \sqrt{x}(1 + sin^2(2\pi/x)) \leq B

and then show that A and B both approach the same value as x approaches 0, thereby squeezing the part in the middle. You need to figure out what to use for A and B. My hint in the previous post might be of help.

 

[realism877]

Would it be the sqr of x?

I assume that I have to plug in 0 for the limit. So by the squeeze theorem, the limit of the function is 0.

Am I on to something?

 

[Mark44]

Would it be the sqr of x?

I assume that I have to plug in 0 for the limit. So by the squeeze theorem, the limit of the function is 0.

Am I on to something?

Well, partly. Where I had A before, you can use sqrt(x). You aren't using the hint, though.

 

[realism877]

I'm lost.

The A won't be negative. Would it?

 

[realism877]

Okay, so it would be 1? But why is it between 0 and 1?

A=0

B=1

 

[Mark44]

No.

Since 0 <= sin²(u) <= 1, what bounds can you place on 1 + sin²(u)? IOW, what's the smallest possible value of 1 + sin²(u)? What's the largest possible value of 1 + sin²(u)?

 

[realism877]

The largest value is 1, and the smallest value is 0.

 

[Mark44]

No for both. The largest value of sin²(u) is 1 and the smallest value is 0, but what happens when you add 1 to sin²(u)?

 

[realism877]

I'm totally lost.

Thanks for helping out though.

 

[realism877]

The best I can do is that I know sin is between -1 and 1. But in this case, it's only coming from the right side. So the limit would be 0.

 

[Mark44]

The best I can do is that I know sin is between -1 and 1.

More precisely, -1 <= sin(u) <= 1 for all real numbers u.
Also, 0 <= sin²(u) <= 1 for all real numbers u.
If you add 1 to all three members of this inequality, what do you get?

I can't do much more without working the problem for you...

But in this case, it's only coming from the right side. So the limit would be 0.

And as I said, the reason for that is that sqrt(x) is not defined for negative numbers.

 

[realism877]

In this case, I would have to add 1 to the right and side and the left hand side to isolate the function?

Correct?

 

[Mark44]

Yes. Since -1 <= sin²(u) <= 1, then
0 <= 1 + sin²(u) <= 2

So, for any real u, 1 + sin²(u) is between 0 and 2.

Now, how can you work this fact into evaluate the limit you're trying to find?

 

[realism877]

From the left hand side, we multiply 1 to sqr(x).

From the right, we multiply 2 to sqr(x).

Once we do that, we calculate the limit, which comes out to 0 for both sides.

My problem is understanding why is 1 remained in the middle of the inequality?

 

[Mark44]

What you wrote is on the cryptic side, so I'll try to help you out. I think this is what you're saying:

\sqrt{x} \leq \sqrt{x}(1 + sin^2(2\pi/x)) \leq 2\sqrt{x}

Since the inequality above is true for all x other than 0, it's true as x approaches 0. Keep in mind which theorem you're using. Hint: it's in the title you picked.

 

[realism877]

That's exactly what I'm saying.

I'm trying to look from an algebraic perspective. I want to know where the 1 came from.

 

[Mark44]

Which 1? The 1 that is a coefficient of the square root on the left? Or the 1 that is in the expression in the middle?

 

[realism877]

The 1 in the expression in the middle.

It is written in the problem, then it is taken out and put back in.

 

[Mark44]

No, it wasn't taken out.

Here's the logic in this problem
For any real number u,
-1 <= sin(u) <= 1
==> 0 <= sin²(u) <= 1
==> 1 <= 1 + sin²(u) <= 2
==> sqrt(x) <= sqrt(x)(1 + sin²(u)) <= 2sqrt(x)

Now, replace u by 2pi/x, and take the limit as x --> 0 from the right.

 

[realism877]

Got it.

Thanks!