<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] You've made some errors in arithmetic.  The solution involves doing both of the things you tried:
$$\begin{align*}
\frac{3-\sqrt{x^2+5}}{\sqrt{2x} - \sqrt{x+2}} &= \frac{(3^2 - (x^2+5))( \sqrt{2x} + \sqrt{x+2})}{(2x - (x+2))(3 + \sqrt{x^2+5})} \\ &= \frac{(4-x^2)(\sqrt{2x} + \sqrt{x+2})}{(x-2)(3 + \sqrt{x^2+5})} \\
&= -(x+2) \frac{\sqrt{2x} + \sqrt{x+2}}{3 + \sqrt{x^2+5}}. \end{align*}$$  Now all expressions are nonzero when $x = 2$, giving $$\lim_{x \to 2} \frac{3-\sqrt{x^2+5}}{\sqrt{2x} - \sqrt{x+2}} = -(2+2) \frac{\sqrt{4} + \sqrt{4}}{3 + \sqrt{9}} = -\frac{8}{3}.$$
As you can see, the reason why you need to do both is because there is a factor of $x-2$ that appears in both the numerator and denominator, and this is what needs to cancel out in order for the limit to be computed.  You did each method by itself but without doing both, that factor cannot be removed.
</s>