Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## Post a reply

Write your message and submit
|
Options

## Topic review (newest first)

noelevans
2012-08-16 08:54:24

Hi Johnathon,

Use the definition:

limf(x)=L     means for each e>0 there is a d>0 such that |x-a|<d ==> |f(x)-L|<e; that is,
x-->a
2                                                                                                   2
lim(x - 1)=3 means for each e>0 there is a d>0 such that |x-2|<d ==> |(x - 1) -3|<e.
x-->2

Now choose d in terms of e.  Example: let d=g(e) but be specific like d=sqrt(e).  You may have to
get creative.  Then you need to "transform"  the |x-2|<g(e) into |x^2-1-3|<e;that is, you need
to show that |x-2|<d implies that |x^2 - 4|<e.

That's the general procedure.  But there are many functions that finding an appropriate d and
doing the "transformation" is nigh unto impossible.  That's why they prove that products, sums,
differences, and quotients (excluding where the denominator may be or approach zero) are
continuous.  This makes all polynomials continuous.  Then x^2-1 is continuous since it is a product
and a difference of the continuous functions x and 1..

Then the nice little theorem that says if a function is continuous at a then the limit as x-->a is f(a).
makes the limit here as x-->2 turn out to be f(2)=2^2 - 1 = 3.

Good luck!

By the way, what would it mean to say that L is NOT the limit of f(x) as x approches a?

Johnathon bresley
2012-08-16 01:10:33

How do i prove this using delta epsilon-
Lim  x^2-1=3
x-->2
(please use d and e for delta and epsilon)