<s>[INST] Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition I want to prove that $$\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$$ but I am not sure if my proof is valid because I did some algebraic manipulation. Can you please verify my proof?

Given $ \epsilon \gt 0 $, Choose $ \delta = \min\{3, \sqrt{2\epsilon}\} $
Suppose $ 0 \lt x - 2 \lt \delta $
Check:
$$\begin{align}
\left|\frac{x - 2}{\sqrt{x^{2} - 4}} + 1 - 1\right| &= \left| \frac{x - 2}{\sqrt{x^{2} - 4}}\right|\\
&= \frac{x - 2}{\sqrt{x^{2} - 4}} \\
&= \frac{x - 2}{\sqrt{x + 2}\sqrt{x - 2}} \\
&= \frac{\left(x - 2\right)^{1}}{\sqrt{x + 2}\cdot\left(x - 2\right)^{0.5}} \\
&= \frac{\sqrt{x - 2}}{\sqrt{x + 2}} 
\end{align}$$
$$0 \lt x - 2 \lt 3\quad\Rightarrow\quad 4 \lt x + 2 \lt 7 
\quad\Rightarrow\quad 2 \lt \sqrt{x + 2} \lt \sqrt{7}$$
$\Rightarrow$
$$\frac{\sqrt{x - 2}}{\sqrt{x + 2}} \lt \frac{\sqrt{\delta}}{2} \le \epsilon\tag*{$\blacksquare$}$$
 [/INST] You should have $\delta<4\varepsilon^2$, so that $\frac{\sqrt\delta}2\leqslant\varepsilon$. So, take $\delta=\min\left\{3,4\varepsilon^2\right\}$.
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