# Epsilon delta to N & M Definition

1. Oct 8, 2008

### prasannaworld

Okay for a simple finite limit: e.g.
lim (3x) = 3
x->1

in the end I say:

"Therefore for every |x - 3| < delta, there exists an epsilon such that |3x-3| < epsilon"

Hence I can make delta really really small and the y bounds of epsilon will constrain the limit.

So lets come to the example I saw in an article
lim (SQRT(x)) = INF
x-INF

Okay so:
x > N - x is greater than any positive integer
Match N with M^2
x > M^2
SQRT(x) > M

Okay so how will I make my statement?

2. Oct 8, 2008

### sutupidmath

so you are trying to prove that:

$$\lim_{x\rightarrow \infty}\sqrt{x}=\infty$$ right?

What we want to prove is that $$\forall N>0, \exists M>0$$ such that whenever

$$x>M, \sqrt{x}>N$$

By observation we have, as you pointed out:$$\sqrt{x}>N=> x>N^2$$

so our statement would be

$$\forall N>0, \exists M=N^2>0$$ such that whenever

$$x>M=N^2=>\sqrt{x}>N$$

3. Oct 8, 2008

### prasannaworld

Yes. That is what I wanted. Still I think the best way for me to get this is convince myself by trying to prove a false limit (I obviously should not be able to...)