$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
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.
 A: Note that $$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$$
The denominator goes to zero, but the numerator goes to $\sqrt{3-\sqrt{3}}$. 
Thus, we know that the limit that you are trying to evaluate does not exist, and the limit goes to $-\infty$ and $\infty$ respectively as $x \to 2$.  
A: $$\lim\limits_{x\to 2^{\pm}}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2} = \left[\frac{\sqrt{3-\sqrt{3}}}{0^{\pm}}\right]=\pm\infty$$
As you can see, the left- and right-side limits are different, so the limit $\lim\limits_{x\to 2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt{3}}}{x-2}$ does not exists.
A: 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.
A: Following StackTD's comment
let $\sqrt{1+\sqrt{x+2}}-\sqrt3=y\implies y\to0$
and $1+\sqrt{x+2}=(y+\sqrt3)^2=y(y+2\sqrt3)+3$
$$x+2=(y^2+2\sqrt3y+2)^2=y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)+4$$
$$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}=\lim_{y\to0}\dfrac y{y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)}=?$$
