<s>[INST] limit problem (with roots) Is it possible to evaluate this limit without graphing or guessing
(ie to replace it by a simpler function)
$$\lim_{x\to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
I tried normalizing by multiplying by the conjugate (both denominator and numerator) didn't work.
 [/INST] $$\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{(\sqrt{6-x}-2)\times (\sqrt{6-x}+2)\times(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)\times(\sqrt{3-x}+1)\times (\sqrt{6-x}+2)}$$
$$=\frac{\left((\sqrt{6-x})^2-4\right)\times(\sqrt{3-x}+1)}{\left((\sqrt{3-x})^2-1\right)\times (\sqrt{6-x}+2)}=\frac{\overbrace{\left((6-x)-4\right)}^{(2-x)}\times(\sqrt{3-x}+1)}{\underbrace{\left((3-x)-1\right)}_{(2-x)}\times (\sqrt{6-x}+2)}$$ 
$$=\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2},~~x\ne 2$$
</s>