Problem states:

Let be a sequence in with , and n.

Prove that = 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 and be sequences of real numbers. If is bounded and = 0,

then = 0."

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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!)