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Prove that $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$ using the epsilon-delta definition.

This is what I have, but I know my delta value is incorrect. My professor said that it was the right path but my delta is incorrect.

Proof: Let $\varepsilon>0$. Choose $\delta$ such that $0<\delta<\min(\varepsilon,1)$. This means that both $\delta<1$ and $\delta<\varepsilon$. Let $x\in\mathbb{R}$ such that $0<|2x-8|<\delta$. Since $\delta<1$, we have

$$\begin{array}{cccccc} &-1 &< & 2x-8 & < & 1\\ \Rightarrow & 7 &<& 2x &<& 9 \\ \Rightarrow & 7/2 & < & x & < & 9/2 \end{array}$$

Since $7/2<x<9/2$,

$$\begin{array}{cccccc} &7/2 & < & x & < & 9/2\\ \Rightarrow & 7 &<& 2x &<& 9 \\ \Rightarrow & 7+7 & < & 2x+7 & < & 9+7 \\ \Rightarrow & \sqrt{14} & < & \sqrt{2x+7} & < & \sqrt{16} \\ \Rightarrow & \sqrt{14} + \sqrt{15} & < & \sqrt{2x+7}+\sqrt{15} & < & \sqrt{16}+\sqrt{15}\\ \Rightarrow & \displaystyle \frac{1}{\sqrt{14} + \sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{2x+7}+\sqrt{15}} & > & \displaystyle \frac{1}{\sqrt{16} + \sqrt{15}}\\ \end{array}$$

This implies $$\left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|< \frac{1}{\sqrt{14} + \sqrt{15}}<1.$$

Therefore,

$$\begin{align*} \left|\sqrt{2x+7}-\sqrt{15}\right| &= \left|\left(\sqrt{2x+7}-\sqrt{15}\right) \cdot \left(\frac{\sqrt{2x+7}+\sqrt{15}}{\sqrt{2x+7}+\sqrt{15}}\right)\right| \\ &= \left|2x+7-15\right| \cdot \left| \frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\ &=\left|2x-8\right|\cdot \left|\frac{1}{\sqrt{2x+7}+\sqrt{15}}\right|\\ &< \delta \cdot 1 \\ &< \varepsilon \cdot 1\\ \end{align*}$$

Thus, $|\sqrt{2x+7}-\sqrt{15}|<\varepsilon$. So, $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$.

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  • $\begingroup$ I know that I need to use $0<|x-4|<\delta$ as this was my original problem, but i cannot figure out the value for delta. $\endgroup$ – Jenna King Apr 2 '17 at 22:17
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    $\begingroup$ well done Jenna King :D $\endgroup$ – SAJW Apr 2 '17 at 22:18
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    $\begingroup$ Best response I've ever seen to "What have you tried?" Plus 1 $\endgroup$ – qbert Apr 2 '17 at 22:20
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    $\begingroup$ I'm not reading the rest of your work, but based on what you did, we should want to require that $0<|x-4|<\min(\varepsilon/2,1/2)$. There may be an easier way to do it. $\endgroup$ – Ted Shifrin Apr 2 '17 at 22:25
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    $\begingroup$ @JennaKing Very nicely explained, your TeX skills are far beyond what mine were at your level. I made a few minor edits: First, props for using \[ \] over $$ $$, but Math SE uses $$ $$. Keep using \[ \] for everything else, though, it's more forward compatible. Second, look up the align and align* environments, they will save you a lot of time formatting arrays. $\endgroup$ – DMcMor Apr 2 '17 at 22:38
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Hint:

$$ \sqrt{2x+7} - \sqrt{15} = \frac {(\sqrt{2x+7} - \sqrt{15} )(\sqrt{2x+7} + \sqrt{15} )}{\sqrt{2x+7} + \sqrt{15} } = \frac{2x-8}{\sqrt{2x+7} + \sqrt{15}} $$

To remove the dependency in $x$ in the denominator, use that square roots are positive: $$ |\sqrt{2x+7} - \sqrt{15}| = \frac {|2x-8|}{\sqrt{2x+7} + \sqrt{15}} \le \frac {|2x-8|}{\sqrt{15}} $$

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Denote by $D$ the domain of $f$, where $f(x) = \sqrt{2x + 7}$.

Let $\varepsilon > 0$. Choose $\delta = (\sqrt{15}/2) \varepsilon$. Let $x \in D$. If $0 < |x-4| < \delta$, then $$\begin{aligned}[t] |f(x) - \sqrt{15}| = \biggl| \dfrac{(f(x) - \sqrt{15}\,)(f(x) + \sqrt{15}\,)}{f(x) + \sqrt{15}}\biggr| &= \dfrac{1}{f(x) + \sqrt{15}} \cdot |(2x+7)-15| \\ &= \dfrac{2}{f(x) + \sqrt{15}} \cdot |x-4| \\ &< \dfrac{2}{\sqrt{15}} \cdot \delta \\ &= \dfrac{2}{\sqrt{15}} \cdot \dfrac{\varepsilon \sqrt{15}}{2} = \varepsilon.\end{aligned} $$

Therefore, $\lim_{x \to 4} \sqrt{2x+7} = \sqrt{15}$.

Remember that if you are trying to prove $\lim_{x \to c} f(x) = L$ using the $\varepsilon$-$\delta$ definition, the structure of your proof should be like so:

"Let $\varepsilon > 0$."
      [A choice for $\delta$ goes here.]
            "Let $x \in \operatorname{dom}(f)$."
                  "Suppose $0 < |x-c| < \delta$."
                  [Proof that $|f(x) - L| < \varepsilon$ goes here.]
"Thus, $\lim_{x \to c} f(x) = L$."

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  • $\begingroup$ What choice should one make for $\delta$? That is the difficult part in $\varepsilon$-$\delta$ proofs. One has to look at the proof of $|f(x) - L| < \varepsilon$ in order to see what $\delta$ can be (there is never exactly one choice, by the way, but there are "good" choices that make the proofs clean). $\endgroup$ – Mark Twain Apr 2 '17 at 22:50
  • $\begingroup$ If I use $\delta = (\sqrt(15)/2)\varepsilon$, what would my other value of $\delta$ be to take the minimum of the two? $\endgroup$ – Jenna King Apr 2 '17 at 22:55
  • $\begingroup$ In some $\varepsilon$-$\delta$ proofs of $\lim_{x \to c} f(x) = L$, you'll simply say, "choose $\delta = \underline{\phantom{xxx}}$." In other such proofs, you might need $|x-c|$ to satisfy more than one condition, such as $|x-c| < A$ and $|x-c| < B$; you can make sure both of these are true by taking $\delta = \min \{ A, B \}$. So whether or not $\delta$ is the minimum of two (or more) numbers really depends on how complicated $|f(x) - L|$ turns out to be. In short: not every $\varepsilon$-$\delta$ proof will require that $\delta$ be taken as the minimum of some numbers. $\endgroup$ – Mark Twain Apr 2 '17 at 23:00

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