<s>[INST] FInd the limit without the l’Hospital’s rule FInd the limit without the l’Hospital’s rule:
$$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x}-\sqrt{x+2}}$$
I have tried multiplying the equation with:
$\frac{\sqrt{2x}+\sqrt{x+2}}{\sqrt{2x}+\sqrt{x+2}}$ and got stuck with 
$\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{2x-(x+2)}$ 
$\implies$ $\frac{(3-\sqrt{x^2+5})(\sqrt{2x}-\sqrt{x+2})}{x-2}$
$\frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}}$ and got stuck with 
$\frac{3^2-(x^2+5)}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$ 
$\implies$ $\frac{14-x^2}{(\sqrt{2x}-\sqrt{x+2})(3+\sqrt{x^2+5})}$
I tried to solve the denominator in the second one (which is the same as the nominator in the first but with different signs) which ended up with: 
$\frac{14-x^2}{3\sqrt{2x}+\sqrt{2x(x^2+5)}-3\sqrt{x+2}-\sqrt{(x+2)(x^2+5)}}$
What can I do after that?
The answer with the l’Hospital’s rule (using Symbolab) $-\frac{8}{3}$
 [/INST] Your method is correct, but there a sign error in the numerator of your first fraction. Doing what you were, it should have been:
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}\color{red}+\sqrt{x+2})}{2x-(x+2)}$$
$$=\frac{(3-\sqrt{x^2+5})(\sqrt{2x}+\sqrt{x+2})}{2x-(x+2)}\cdot\color{blue}{\frac{(3+\sqrt{x^2+5})}{(3+\sqrt{x^2+5})}}$$
You can do it fairly easily now...
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