# Thread: Intro to Real Analysis help: Proving limit is zero.

1. ## Intro to Real Analysis help: Proving limit is zero.

Problem states:

Let be a sequence in with , and n.
Prove that = 0.

---------------------------------------------
Note: The following had to be proven previously, so I assume it will help in completing the above proof: "Let and be sequences of real numbers. If is bounded and = 0,
then = 0."

----------------------------------------------

My ideas so far:
Since {$\displaystyle x_n$} → 0, I know $\displaystyle x_n$ is bounded, and by the theorem if I can show sin (1/$\displaystyle x_n$) → 0 then I'm done.

Or, since $\displaystyle x_n$ → 0, if I can show sin (1/$\displaystyle x_n$) is bounded, then I'm done.

My question now is, which of the two approaches is correct, and how would I go about doing that?

I thought I could do this:
I know sin (n) is bounded by [-1, 1] so,
-1 ≤ sin (n) ≤ 1 so dividing all by n gives,
-1/n ≤ sin (n)/n ≤ 1/n
I know both -1/n and 1/n converge to zero, so by squeeze theorem,
sin (n) / n converges to 0 too. But my classmate pointed out that
sin (n) / n is not the same as sin (1/$\displaystyle x_n$). So I can't do that. :| Therefore I am stuck.

Any help, suggestions, corrections, ideas are greatly appreciated.
Thank you for your time!

(Don't forget DST ends this weekend!)

2. $\displaystyle \left( {\forall x} \right)\left[ {\left| {\sin (x)} \right| \leqslant 1} \right]$
So the sequence $\displaystyle \left( {\sin \left( {\frac{1}{x_n}} \right)} \right)$ is bounded.

3. Originally Posted by Plato
$\displaystyle \left( {\forall x} \right)\left[ {\left| {\sin (x)} \right| \leqslant 1} \right]$
So the sequence $\displaystyle \left( {\sin \left( {\frac{1}{x_n}} \right)} \right)$ is bounded.
How can I get the x in $\displaystyle \sin (x)$ to be 1/$\displaystyle x_n$ to show that sin (1/$\displaystyle x_n$) is bounded? I'm not quite "seeing" how to get to that conclusion right away since sin(x) is bounded by 1.

4. This is true $\displaystyle \left[ {\left| {\sin (?)} \right| \leqslant 1} \right]$ no matter what you choose to put in place of the ?.

5. Originally Posted by Plato
This is true $\displaystyle \left[ {\left| {\sin (?)} \right| \leqslant 1} \right]$ no matter what you choose to put in place of the ?.
Okay, thanks! I got it now. So that makes the proof pretty short then, or would I have to prove somehow that sin (x) is bounded?