Q.$\displaystyle\lim_{x\to 2}\left[\frac{\sqrt{1+\sqrt{2+x}}-\sqrt3}{x-2}\right]$

Multiply by the conjugate $\sqrt{1+\sqrt{2+x}}+\sqrt3$: the numerator becomes $\left(1+\sqrt{2+x}\right)-3=\sqrt{2+x}-2$. Now rationalize this again, multiplying (top and bottom of this inner piece) by $\sqrt{2+x}+2$: it becomes $(2+x)-4=x-2$. So overall, $\dfrac{\sqrt{1+\sqrt{2+x}}-\sqrt3}{x-2}=\dfrac{1}{(\sqrt{2+x}+2)\left(\sqrt{1+\sqrt{2+x}}+\sqrt3\right)}$. At $x=2$: $\sqrt{2+x}+2=4$ and …

Apply the rationalizing (conjugate) trick twice in succession — first to clear the outer square root, then to clear the i …

Stopping after only one rationalization step, which still leaves a $0/0$ form because the inner radi …