Prove that $\lim\limits_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$.

Question

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, $\displaystyle \lim_{x\rightarrow 4} \sqrt{2x+7} = \sqrt{15}$.

Answer 1

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: $$ |\sqrt{2x+7} - \sqrt{15}| = \frac {|2x-8|}{\sqrt{2x+7} + \sqrt{15}} \le \frac {|2x-8|}{\sqrt{15}} $$

Answer 2

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, $\lim_{x \to c} f(x) = L$."