Answer
$\lim\limits_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{1}{2}$
Work Step by Step
$A=\lim\limits_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$
$A=\lim\limits_{x\to2}\frac{B}{C}$
Multiply both numerator and denominator by $(\sqrt{6-x}+2)(\sqrt{3-x}+1)$, we have
- In the numerator:
$B=(\sqrt{6-x}-2)(\sqrt{6-x}+2)(\sqrt{3-x}+1)$
$B=[(6-x)-4](\sqrt{3-x}+1)$
(apply $(a-b)(a+b)=a^2-b^2$)
$B=(2-x)(\sqrt{3-x}+1)$
- In the denominator:
$C=(\sqrt{3-x}-1)(\sqrt{3-x}+1)(\sqrt{6-x}+2)$
$C=[(3-x)-1](\sqrt{6-x}+2)$
(apply $(a-b)(a+b)=a^2-b^2$)
$C=(2-x)(\sqrt{6-x}+2)$
Therefore,
$A=\lim\limits_{x\to2}\frac{(2-x)(\sqrt{3-x}+1)}{(2-x)(\sqrt{6-x}+2)}$
$A=\lim\limits_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$
$A=\frac{\sqrt{3-2}+1}{\sqrt{6-2}+2}$
$A=\frac{1}{2}$
