A delta epsilon proof, using graph to find delta > 0

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Oneiromancy

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Sep 28, 2007
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I just need help on how to treat the absolute value sign.

I have this:

abs { (2 / sqrt (x + 1) + 2) * (x - 3) } < epsilon

What I really want is to move blah blah * (x - 3) to the other side so I can solve for delta (it tells me what epsilon is in the book). How do I do that? There might be another way, I'm new at this.
 
Re: A delta epsilon proof.

Are you trying to prove a limit? Can you post the problem in its entirety?
 
Re: A delta epsilon proof.

Sure:

Use the graphs to find a delta > 0 such that for all x, 0 < |x - a| < delta ==> |f(x) - L| < epsilon

10. (picture of graph showing delta and epsilon, but i still have to show work/proof other than pointing to graph)

f(x) = 2*sqrt ( x+1 )
a = 3
L = 4
epsilon = 0.2

The delta should eventually be 0.39 for the given epsilon, 0.2.
 
Re: A delta epsilon proof.

So just using the definition of a limit:

Given e > 0, there exists d > 0 such that whenever 0 < |x - 3| < d, then it follows that |2sqrt{x + 1} - 4| < 0.2

Now, looking at the latter inequality, we want to get rid of the absolute value sign. So:

\(\displaystyle | 2 \sqrt{x+1}-4 < 0.2\)

\(\displaystyle -0.2 < 2\sqrt{x+1}-4<0.2\)

Can you carry on from here?
 
Re: A delta epsilon proof.

Ohhh ya thanks.

So basically I'm going to add 4 to everything, square everything, etc. eventually I'll get x - 3 in which case I'll be able to convert it to absolute value (since both sides equal).

So what will that tell me once I do all that?
 
Re: A delta epsilon proof.

What you have written is hard for me to read.
If it is
\(\displaystyle abs\left({\frac{2}{{\sqrt {x + 1} + 2}}\left( {x - 3} \right)}\right) < \varepsilon\)
then you can write
\(\displaystyle \frac{{4\left( {x - 3} \right)^2 }}{{\left( {\sqrt {x + 1} + 2} \right)^2 }} < \varepsilon ^2\)
 
Re: A delta epsilon proof.

Well eventually, you'll solve get:

|x - 3| < (some number) right?

Well, remember at the beginning we said that:

Given e = 0.2, there exists some d > 0 such that whenever 0 < |x - 3| < d then it NECESSARILY FOLLOWS that |2sqrt{x + 1} - 4| < 0.2.

You basically worked backwards from 2sqrt{x+1} - 4 < 0.2 and got |x - 3| < (some number) which is what d should be.
 
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