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prasannaworld
Oct8-08, 03:30 AM
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?

sutupidmath
Oct8-08, 03:38 AM
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

prasannaworld
Oct8-08, 03:43 AM
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...)