<s>[INST] How to define $\delta$ to prove $\lim\limits_{x \to 9} \sqrt{x-5} = 2$ 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.
 [/INST] 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 $\delta=\min(2\epsilon,4).$
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