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A Limit Proof Using Estimation

Date: 06/01/98 at 08:33:39
From: mostyn
Subject: Limit

How do I show that the limit of x^2 as x->(-2) is 4, using the
delta-epsilon definition of a limit?  


Date: 06/01/98 at 11:02:37
From: Doctor Rob
Subject: Re: Limit

The definition says that for any epsilon > 0, no matter how small, you 
can find a delta > 0 such that:

   |x-(-2)| = |x+2| < delta implies that |x^2-4| < epsilon

The idea is to start with what you want to show, that 
|x^2-4| < epsilon, and to manipulate this until you can get it
into the form |x+2| < some expression in epsilon. Then picking
delta to be this expression in epsilon will do, and the proof
is to work backwards through the steps of the manipulation.

In this case:

   |x^2-4| < epsilon
    <==> |(x+2)^2 - 4*(x+2)| < epsilon
    <==  |x+2|^2 + 4*|x+2| < epsilon (by the triangle inequality)
    <==> |x+2|^2 + 4*|x+2| - epsilon < 0

Now you can use the Quadratic Formula to solve for |x+2|, and thus 
find an upper bound on |x+2| in terms of epsilon. That will be what 
you choose for delta.

One tricky part is that each step needs an implication arrow in one 
direction (<==) but not necessarily in the other.

Another tricky part is that there isn't necessarily a unique answer. 
In this case, you could have proceeded like this instead:

   |x^2-4| < epsilon
    <==> |(x+2)*(x-2)| < epsilon
    <==> |x+2|*|x-2| < epsilon
    <==  |x+2|*(4 + |x+2|) < epsilon (by the triangle inequality)
    <==  5*|x+2| < epsilon (provided |x+2| <= 1)

and so on. At the end, you pick delta to be the minimum of 1 and the 
expression involving epsilon, and this ensures the "provided ..." 

Other expressions for delta in terms of epsilon may also work.

-Doctor Rob,  The Math Forum
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Associated Topics:
High School Analysis

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