why is $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to 2 $ \sqrt x - \sqrt2 = \sqrt{(x-2)} $ when x tends to $ 2^+ $
i have this problem
lim when x tends to $ 2^+ $
$ \frac{\sqrt x -\sqrt2 +\sqrt{x-2}}{\sqrt{x^2-4}} $
i know i must group $ \sqrt x - \sqrt 2 $ into $ \sqrt{x-2} $ only because that is true when x tends to 2 but then i separate the sums and simplify but the calculator gives me another result here's what i did
$  \frac{\sqrt{x-2} +\sqrt{x-2}}{\sqrt{x^2 -4}} $
$\frac{\sqrt{x-2}}{\sqrt{x^2 -4}} + \frac{\sqrt{x-2}}{\sqrt{x^2 -4}}  $
$ \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} + \frac{\sqrt{x-2}}{\sqrt{x-2}\sqrt{x+2}} $
simplifying
$ \frac{1}{\sqrt{x+2}} + \frac{1}{\sqrt{x+2}} $
this is when x tends to $ 2^+ $
$ \frac{2}{\sqrt{4}} = 1  $
but the calculator gets me $ \frac{\sqrt{2}}{2} $
where did i go wrong?

You need to evaluate $$\lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^{2} - 4}}$$ and you are not supposed to replace $\sqrt{x} - \sqrt{2}$ by $\sqrt{x - 2}$. This is not allowed through any rule of algebra/calculus. What we do here is not difficult, but a bit of careful and simple algebraical manipulation:
$\displaystyle \begin{aligned}\lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2} + \sqrt{x - 2}}{\sqrt{x^{2} - 4}} &= \lim_{x \to 2^{+}}\frac{\sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}} + \frac{\sqrt{x - 2}}{\sqrt{x^{2} - 4}}\\
&= \lim_{x \to 2^{+}}\frac{(\sqrt{x} - \sqrt{2})(\sqrt{x} + \sqrt{2})}{(\sqrt{x} + \sqrt{2})\sqrt{x^{2} - 4}} + \frac{\sqrt{x - 2}}{\sqrt{(x - 2)(x + 2)}}\\
&= \lim_{x \to 2^{+}}\frac{x - 2}{(\sqrt{x} + \sqrt{2})\sqrt{(x - 2)(x + 2)}} + \frac{1}{\sqrt{x + 2}}\\
&= \lim_{x \to 2^{+}}\frac{\sqrt{x - 2}}{(\sqrt{x} + \sqrt{2})\sqrt{x + 2}} + \frac{1}{\sqrt{x + 2}} = 0 + \frac{1}{\sqrt{2 + 2}} = \frac{1}{2}\end{aligned}$
