Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Need Help proving limit

Prove limx->2  x^3 = 8
let epsilon > 0 be given for x.
|f(x)-L] = |x^3-8|
is < epsilon if o<|x-2|<delta
then |f(x)-L| = |x^3-8| < epsilon

Assuming im right so far what do I do now? Also, can some1 plz explain to me how to relation epsilon to delta, im not sure how and I dont think I did it right on this question.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Need Help proving limit

Well, the tricky part is the |x[sup]3[/sup]−8|, isn’t it? Here’s one trick you could use.

Note that

.

The curve

, which is minimum at x = −1, is strictly increasing for x > −1. Now you have

. So if we make sure δ is not more than 1, x will be less than 3 (and greater than 1), so x[sup]2[/sup]+2x+4 will be guaranteed to be less than 19.

So if 0 < |x−2| < δ and δ ≤ 1, we have

. We also want 19δ ≤ ε, i.e. δ ≤ ε⁄19. Hence

Last edited by JaneFairfax (2007-09-17 23:52:11)