What is sin x when x tends to infinity?

This question popped up in my head. What is sin x when x tends to infinity? Since sine is a preiodic functions which repeats itself, is the answer 1, -1 or 0 or something else altogether?
 
R

rachmaninoff

It is undefined; there is no limit! Sin[x] does not converge to any value as x increases - it maintains its periodic character. Limits only apply when something converges to something somewhere.

A proof is readily constructed from the def. of limits at [tex]\mbox{$\infty$}[/tex]. Choose a small [tex]\mbox{$\epsilon$}[/tex] ("1"is small enough) and show that for any x0, no matter how large, there exists an x>x0 such that

[tex]\|f(x)-f(x_0)\|\geq\epsilon[/tex].​

Plug in [tex]\mbox{$\epsilon=1$}[/tex] and [tex]\mbox{$x=x_0+\frac{\pi}{2}$}[/tex].

What you show is that is not convergent, thus there is no limit.

Hey, does anyone know a better way to do inline LaTeX here?
 
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dextercioby

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It looks good to me until [itex]\geq[/itex].

Daniel.
 
dextercioby said:
It looks good to me until [itex]\geq[/itex].

Daniel.
What's wrong with [itex]\geq[/itex]? You want to show that [itex]\|f(x)-f(x_0)\|<\epsilon[/itex] does not hold for all [itex]x>x_0[/itex], so you find an [itex]x[/itex] where the relation is [itex]\geq[/itex] instead of [itex]<[/itex].
 

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