How to define $\delta$ to prove $\lim\limits_{x \to 9} \sqrt{x-5} = 2$

Question

When tackling this problem, parting from the assumption that $\lvert \sqrt{x-5} - 2\rvert \lt \epsilon$, I arrived through arithmetical manipulation at $$\frac{\lvert x-9\rvert}{\lvert \sqrt{x-5} + 2\rvert} \lt \epsilon,$$ which has the $\lvert x-9\rvert$ needed to prove that for a certain $\delta$, $0<|x-9|<\delta \Rightarrow |\sqrt{x-5} + 2|<\epsilon$. However, I don't know how to eliminate the $x$ in ${\lvert \sqrt{x-5} + 2\rvert}$, such that I can define $\delta$ solely based on $\epsilon$. I would appreciate any help.

Answer 1

Assuming $|x-9|<\delta$ we have $$ \left| \sqrt{x-5} - 2 \right| = \left| \frac{(\sqrt{x-5} - 2)(\sqrt{x-5} + 2)}{\sqrt{x-5} + 2} \right| = \left| \frac{x-9}{\sqrt{x-5} + 2} \right| < \frac{\delta}{\sqrt{x-5} + 2} . $$ Now note that $\frac{1}{\sqrt{x-5} + 2} < \frac{1}{2}$ as long as $x-5>0.$ The latter can be achieved by making sure that $|x-9|<4.$ Therefore we will require $\delta<4.$ Then we can just take $\delta\leq 2\epsilon$ and we can continue: $$ \cdots < \frac{\delta}{2} \leq \epsilon. $$

Thus, given $\epsilon>0$, take $\delta=\min(2\epsilon,4).$

Answer 2

Hints: For this particular problem , for convenient first we'll calculate the value of $\delta$ ,

We have $$|\sqrt{x-5}-2|<\epsilon$$

$$-\epsilon<\sqrt{x-5}-2<\epsilon$$

$$(2-\epsilon)^2-4<x-5<(2+\epsilon)^2-4$$.

Now choose $\delta=\text{min}\{(2-\epsilon)^2-4,(2+\epsilon)^2-4$}.