# Limit of a function help please

1. Oct 30, 2012

### Questions999

Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε

I know how to keep this going,but is it correct until now?

2. Oct 30, 2012

### Staff: Mentor

The right side should be |[(-1 )^(n-1)]/n^2| < ε
|(-1)^(n - 1)| = 1, so all you need to do is find n so that 1/n^2 < ε.
Put parentheses around your exponent. What you wrote, (-1)^n - 1 would be interpreted as
$$(-1)^n - 1$$

3. Oct 30, 2012

### Questions999

Thank you so much! :)