Intro to Real Analysis help: Proving limit is zero.

**Problem states:**

Let http://www.cramster.com/Answer-Board...9375003005.gif be a sequence in http://www.cramster.com/Answer-Board...1250009648.gif with http://www.cramster.com/Answer-Board...9062501798.gif, and http://www.cramster.com/Answer-Board...5312508866.gif http://www.cramster.com/Answer-Board...0000002684.gif n.

Prove that http://www.cramster.com/Answer-Board...4687509666.gif http://www.cramster.com/Answer-Board...2187507217.gif= 0.

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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,

then http://www.cramster.com/Answer-Board...4687509666.gif http://www.cramster.com/Answer-Board...2500008978.gif= 0."

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My ideas so far:

Since { } → 0, I know is bounded, and by the theorem if I can show sin (1/ ) → 0 then I'm done.

Or, since → 0, if I can show sin (1/ ) 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/ ). 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!)