1. ## Delta Epsilon proof

So I understand the general idea behind a delta epsilon proof but for some reason this one has me stumped. Any help would be appreciated! Thanks!

Attached I included the question and what I have so far. I am actually stumped on how to continue from there. Thank you!

2. ## Re: Delta Epsilon proof

If |x+ 3|< 1 then -1< x+ 3< 1. So 3< x+ 7< 5 and $\displaystyle \sqrt{3}< \sqrt{x+ 7}< \sqrt{5}$. $\displaystyle 2+ \sqrt{3}< \sqrt{x+ 7}+ 2< 2+ \sqrt{5}$. Then $\displaystyle \frac{1}{2+ \sqrt{5}}< \frac{1}{\sqrt{x+ 7}+ 2}< \frac{1}{2+ \sqrt{3}}$.

Replace the $\displaystyle \frac{1}{\sqrt{x+ 7}+ 2}$ with othe upperbound $\displaystyle \frac{1}{2+ \sqrt{3}}$.

3. ## Re: Delta Epsilon proof

let $\epsilon > 0$

$|\sqrt{x+7}-2| < \epsilon$

$-\epsilon < \sqrt{x+7}-2 < \epsilon$

$2- \epsilon < \sqrt{x+7} < 2+\epsilon$

$4 - 4\epsilon + \epsilon^2 < x+7 < 4 + 4\epsilon + \epsilon^2$

$\epsilon^2 - 4\epsilon < x+3 < \epsilon^2 + 4\epsilon$

let $\delta = 3\epsilon$

$|x+3| < \delta = 3\epsilon$

$-3\epsilon < x+3 < 3 \epsilon \overset{\text{small }\epsilon}{\Rightarrow} \epsilon^2 - 4\epsilon < x+3 < \epsilon^2 + 4\epsilon \Rightarrow |\sqrt{x+7}-2| < \epsilon$

so

$\forall \epsilon > 0,~\exists \delta = 3\epsilon > 0 \ni |x+3|<\delta \Rightarrow |\sqrt{x+7}-2| < \epsilon$

This is the epsilon-delta limit definition of $\displaystyle \lim_{x\to -3} \sqrt{x+7} = 2$