# What is sin x when x tends to infinity?

1. Jun 4, 2005

### 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?

2. Jun 4, 2005

### 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?

Last edited by a moderator: Jun 5, 2005
3. Jun 5, 2005

### dextercioby

It looks good to me until $\geq$.

Daniel.

4. Jun 5, 2005

### master_coda

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 $<$.