Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition

Question

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?

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Given $ \epsilon \gt 0 $, Choose $ \delta = \min\{3, \sqrt{2\epsilon}\} $

Suppose $ 0 \lt x - 2 \lt \delta $

Check:

$$\begin{align} \left|\frac{x - 2}{\sqrt{x^{2} - 4}} + 1 - 1\right| &= \left| \frac{x - 2}{\sqrt{x^{2} - 4}}\right|\\ &= \frac{x - 2}{\sqrt{x^{2} - 4}} \\ &= \frac{x - 2}{\sqrt{x + 2}\sqrt{x - 2}} \\ &= \frac{\left(x - 2\right)^{1}}{\sqrt{x + 2}\cdot\left(x - 2\right)^{0.5}} \\ &= \frac{\sqrt{x - 2}}{\sqrt{x + 2}} \end{align}$$

$$0 \lt x - 2 \lt 3\quad\Rightarrow\quad 4 \lt x + 2 \lt 7 \quad\Rightarrow\quad 2 \lt \sqrt{x + 2} \lt \sqrt{7}$$

$\Rightarrow$

$$\frac{\sqrt{x - 2}}{\sqrt{x + 2}} \lt \frac{\sqrt{\delta}}{2} \le \epsilon\tag*{$\blacksquare$}$$

Answer 1

You should have $\delta<4\varepsilon^2$, so that $\frac{\sqrt\delta}2\leqslant\varepsilon$. So, take $\delta=\min\left\{3,4\varepsilon^2\right\}$.

Answer 2

HINT

A slightly different approach. Suppose that $0 < x - 2 < \delta_{\varepsilon}$. Then it results that \begin{align*} \left|\frac{x - 2}{\sqrt{x^{2} - 4}}\right| = \frac{\sqrt{(x - 2)^{2}}}{\sqrt{(x - 2)(x + 2)}} = \frac{\sqrt{x - 2}}{\sqrt{x + 2}} < \frac{\sqrt{x - 2}}{2} < \frac{\sqrt{\delta_{\varepsilon}}}{2} := \varepsilon \end{align*}

Can you take it from here?