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?

The only way you can do is to make $\sqrt{x}-\sqrt{2}$ into $x-2$ by $$\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x}+\sqrt{2}}=\frac{x-2}{\sqrt{x}+\sqrt{2}}$$ So the main fraction can be changed to $$\sqrt{\frac{x-2}{x+2}}\times\frac{1}{\sqrt{x}+\sqrt{2}}+\frac{1}{\sqrt{x+2}},~~x\neq 2$$
