- #1

fishturtle1

- 394

- 82

## Homework Statement

Find and prove ##\operatorname{lim} \frac {1}{n^2}##.

## Homework Equations

In the textbook, we assume that the limit is going to infinity without writing it.

If L is the limit, we have for all ##\epsilon > 0##, there exists ##N## such that ##n \epsilon \mathbb{Z}## and ##n > N## implies ##\vert \frac{1}{n^2} - L \vert < \epsilon##.

## The Attempt at a Solution

This is an example in the book. I can follow how to solve for N if we guess ##\operatorname{lim} \frac {1}{n^2} = 0##.

We just have ##\vert \frac {1}{n^2} - 0 \vert < \frac {1}{n^2} < \epsilon##.

So ##n > \frac {1}{\sqrt{\epsilon}}##. Then we work backwards in the actual proof.

I am confused, if we guess ##\operatorname{lim} \frac {1}{n^2} = 4## where 4 is just some arbitrary number... then we have,

##\vert \frac {1}{n^2} - 4 \vert = \frac {1}{n^2} - 4 < \frac {1}{n^2} < \epsilon##. So, ##n^2 > \frac {1}{\epsilon}##. So ##n > \frac {1}{\sqrt{\epsilon}}##.

So..

Proof: Let ##\epsilon > 0## and choose ##N = \frac {1}{\sqrt{\epsilon}}##. Suppose ##n > N## and ##n## is an integer. Then ##n > \frac {1}{\sqrt{\epsilon}}##. This implies ##n^2 > \frac {1}{\epsilon}## which implies ##\frac {1}{n^2} < \epsilon##. So ##\epsilon > \frac {1}{n^2} > \frac {1}{n^2} - 4 = \vert \frac {1}{n^2} - 4 \vert##. This proves that ##\operatorname{lim} \frac{1}{n^2} = 4.##

But limits are unique... but I'm not sure what I'm doing wrong or why this doesn't work?