# Epsilon delta proof for $\lim_{x\rightarrow 2} \sqrt{x+7} = 3$

I have to proof this limit. $$\lim_{x\rightarrow 2} \sqrt{x+7} = 3$$ I wrote this:

Based on definition, I have to show that to ε > 0 ∃δ> 0, such that: |(√(x+7)) - 3| < ε whenever 0 < |x - 2| < δ.

I will choose δ, looking in inequality:

|(√(x+7)) - 3| < ε |√(x+7) - 3| < ε

But, I don't have an idea what to do after this, because I need some constant for to compare with my δ pulling out of the inequality.

Let $|x-2|<1$, then $1<x<3;$

$|\sqrt{x+7}-3|=$

$|\sqrt{x+7}-3|\dfrac{|\sqrt{x+7}+3|}{|\sqrt{x+7}+3|}=$

$|x-2|\dfrac{1}{|\sqrt{x+7}+3|}\lt$

$\dfrac{|x-2|}{\sqrt{8}}$

$\epsilon >0$ be given.

Choose $\delta < \min (1,√8\epsilon)$

Then

$|\sqrt{x+7}-3| \lt \dfrac{|x-2|}{\sqrt{8}}<$

$\delta/√8 \lt \epsilon$.

• Thank a lot! Now I after your answers I can to continue my studies. Apr 8, 2018 at 13:33
• Juliane.Welcome.With a little practice you will find many epsilon-delta problems quite similar:). Apr 8, 2018 at 14:06

Observe, $$| \sqrt{x+7} - 3 | < \varepsilon \qquad \qquad 0 < | x- 2 | < \delta$$

$$- \varepsilon + 3 < | \sqrt{x + 7}| < \varepsilon + 3$$ $$(- \varepsilon + 3)^2 - 9 < x - 2 < (\varepsilon +3)^2 -9$$ $$\implies \delta = min(\quad (-\varepsilon + 3 )^2-9 ,\quad (\varepsilon + 3)^2 -9\quad )$$

• You probably intended to use a varepsilon on that last one. Apr 8, 2018 at 2:25