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

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

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