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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.

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2 Answers 2

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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}}< $

$\delta/√8 \lt \epsilon$.

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  • $\begingroup$ Thank a lot! Now I after your answers I can to continue my studies. $\endgroup$
    – Ju Grigori
    Apr 8, 2018 at 13:33
  • $\begingroup$ Juliane.Welcome.With a little practice you will find many epsilon-delta problems quite similar:). $\endgroup$ Apr 8, 2018 at 14:06
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Observe, $$ | \sqrt{x+7} - 3 | < \varepsilon \qquad \qquad 0 < | x- 2 | < \delta $$

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

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    $\begingroup$ You probably intended to use a varepsilon on that last one. $\endgroup$
    – Brian Tung
    Apr 8, 2018 at 2:25

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