What is sin x when x tends to infinity?

[misogynisticfeminist]

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?

[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 \mbox{$\infty$}. Choose a small \mbox{$\epsilon$} ("1"is small enough) and show that for any x0, no matter how large, there exists an x>x0 such that

\|f(x)-f(x_0)\|\geq\epsilon.

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

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?

[dextercioby]

It looks good to me until \geq.

Daniel.

[master_coda]

It looks good to me until \geq.

Daniel.

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