- #1

schlynn

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## Homework Statement

Prove that lim x->3 of (x[tex]^{2}[/tex]+x-5=7

## Homework Equations

0<x-c<[tex]\delta[/tex] and |f(x)-L|<[tex]\epsilon[/tex]

## The Attempt at a Solution

The preliminary analysis.

The first equation in the relevant equations becomes

0<x-3<[tex]\delta[/tex]

And the second equation becomes

|(x[tex]^{2}[/tex]+x-5)-7|<[tex]\epsilon[/tex]

=|x[tex]^{2}[/tex]+x-12|<[tex]\epsilon[/tex]

=|(x+4)(x-3)|<[tex]\epsilon[/tex]

Now this is were it gets confusing. The book says "The factor |x-3| can be made as small as we wish, and we know that |x+4| will be about 7." Then it says that this comes from setting |x-3|[tex]\leq[/tex]1

Then from that I would get x[tex]\leq[/tex]4

So what do I do then? The book continues on with

|x+4|=|x-3+7|

[tex]\leq[/tex]|x-3|+|7| (Triangle Inequality)

<1+7=8

So this means that 8 is the bound. But my problem is what factor do I set [tex]\leq[/tex]1? Also, what's going on in the steps immediately above? I can do the delta epsilon proofs that don't require this step, but I don't understand this. I have seen it explained graphically it didn't really help.