$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sqrt{x+2}}}{(x-2)\sqrt{1+\sqrt{x+2}+\sqrt3}}$
I am stuck here.

Following my comment, the question would make "more sense" (seeing that you know the method of multiplying with a conjugate expression) if the numerator was $\sqrt{1+\sqrt{x+2}}-\sqrt3$. 
If you did copy the question correctly, then it might have been a bit of a trick question since you don't need that method; it is not an indeterminate form. Other answers (by S.C.B. and Jaroslaw Matlak) explain the way to go.
If the numerator is as I suggested, then your idea of multiplying with the conjugate expression is fine and you can apply that trick twice in order to cancel out a common factor $\color{red}{x-2}$:
$$\begin{align}\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}
& = \lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}\color{blue}{\frac{\sqrt{1+\sqrt{x+2}}+\sqrt3}{\sqrt{1+\sqrt{x+2}}+\sqrt3}} \\[7pt]
& = \lim_{x\to2}\frac{1+\sqrt{x+2}-3}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)} \\[7pt]
& = \lim_{x\to2}\frac{\sqrt{x+2}-2}{\left(x-2\right)\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)}\color{blue}{\frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}} \\[7pt]
& = \lim_{x\to2}\frac{\color{red}{x-2}}{\color{red}{\left(x-2\right)}\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)} \\[7pt]
& = \lim_{x\to2}\frac{1}{\left(\sqrt{1+\sqrt{x+2}}+\sqrt3\right)\left(\sqrt{x+2}+2\right)}= \frac{1}{8\sqrt{3}} \\[7pt]
 \end{align}$$
Perhaps this still helps.
