Intro to Real Analysis help: Proving limit is zero.

• Nov 1st 2008, 09:05 AM
ilikedmath
Intro to Real Analysis help: Proving limit is zero.
Problem states:

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Note: The following had to be proven previously, so I assume it will help in completing the above proof: "Let http://www.cramster.com/Answer-Board...6875004376.gifand http://www.cramster.com/Answer-Board...2187502633.gif be sequences of real numbers. If http://www.cramster.com/Answer-Board...8437506419.gif is bounded and http://www.cramster.com/Answer-Board...4687509666.gif http://www.cramster.com/Answer-Board...7812505844.gif= 0,

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

Or, since $x_n$ → 0, if I can show sin (1/ $x_n$) 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/ $x_n$). So I can't do that. :| Therefore I am stuck.

Any help, suggestions, corrections, ideas are greatly appreciated.

(Don't forget DST ends this weekend!)
• Nov 1st 2008, 09:22 AM
Plato
$\left( {\forall x} \right)\left[ {\left| {\sin (x)} \right| \leqslant 1} \right]$
So the sequence $\left( {\sin \left( {\frac{1}{x_n}} \right)} \right)$ is bounded.
• Nov 1st 2008, 10:24 AM
ilikedmath
Quote:

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

How can I get the x in $\sin (x)$ to be 1/ $x_n$ to show that sin (1/ $x_n$) is bounded? I'm not quite "seeing" how to get to that conclusion right away since sin(x) is bounded by 1. (Thinking)
• Nov 1st 2008, 10:32 AM
Plato
This is true $\left[ {\left| {\sin (?)} \right| \leqslant 1} \right]$ no matter what you choose to put in place of the ?.
• Nov 1st 2008, 11:50 AM
ilikedmath
Quote:

Originally Posted by Plato
This is true $\left[ {\left| {\sin (?)} \right| \leqslant 1} \right]$ 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?