# Quick question about the range of a sine function

1. Sep 26, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
I am doing a proof about limits, and I need to know the range that sin(n^3-2n)/n is within for all n.

2. Relevant equations

3. The attempt at a solution

I know that sin(n) is in between -1 and 1 for all n, so sin(n)/n would be in between -1/n and 1/n for all n. However, I don't know how to manipulate the range if the inside of the sin(n) function is changed. Any help would be great!!!

2. Sep 26, 2009

### lurflurf

Is this for n->infinity?
if so
sin(n^3-2n) is between -1 and 1
is what you need
then consider 1/n

3. Sep 26, 2009

### dancergirlie

yes this is for n as n approaches infinity. Would it be between -(n^2-2)/(n^3-2n) and (n^2-2)/(n^3-2n)?

4. Sep 26, 2009

### dancergirlie

nevermind i misread your comment, i see that it is in between -1/n and 1/n. Thanks for the help!

5. Sep 26, 2009

### Bohrok

I would find the limit as n->0 and define the function at n=0 to make it continuous. Then find its local maxes and mins from its derivative.