# A delta-epsilon proof.

1. Jan 18, 2008

### Oneiromancy

I'm just kind of confused on how to handle the absolute value along with the inequality. Bad algebra skills. :(

0 < |x - x_0| < $$\delta$$ $$\Rightarrow$$ | f (x) - L | < $$\epsilon$$

I'm given: f(x) = 2$$\sqrt{x + 1}$$, x$$_{0}$$ = 3, L = 4, $$\epsilon$$ = 0.2

The attempt at a solution

Nevermind, I tried latex and it messed everything up. Sorry.

Last edited: Jan 18, 2008
2. Jan 19, 2008

### arildno

Hi, there!

It is best to forget about the epsilon at first, and focus on bounding the difference in terms of delta.

So, we are to bound:
$$|\sqrt{2x+1}-\sqrt{2x_{0}+1}|=|\frac{2x+1-(2x_{0}+1)}{\sqrt{2x+1}+\sqrt{2x_{0}+1}}|=2\frac{|x-x_{0}|}{|\sqrt{2x+1}+\sqrt{2x_{0}+1}|}<2\frac{\delta}{\sqrt{2x_{0}+1}}$$

You may use this to determine a delta that surely will work for som particular epsilon, say:
$$\delta=\frac{\epsilon}{2}\sqrt{2x_{0}+1}$$

Edit:
Seems I used the wrong function, but the technique is similar for your case. Try it out.

Last edited: Jan 19, 2008
3. Jan 19, 2008

### Oneiromancy

Why did you do 2x_0 + 1?

4. Jan 19, 2008

### arildno

Did you see my edit?