Delta-Epsilon Proof for f(x) = 2√(x + 1) at x = 3 with L = 4 and ε = 0.2

[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.

 

[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}|}&lt;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.

 

[Oneiromancy]

Why did you do 2x_0 + 1?

 

[arildno]

Did you see my edit?

Follow a similar procedure with |2*sqrt(x+1)-2*sqrt(x0+1)| instead.

Don't bother to use digit-written numbers(like using 3 instead of x0) before the end.