Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{1}{2}$
$A=\lim\limits_{x\to2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ $A=\lim\limits_{x\to2}\frac{B}{C}$ 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, $A=\lim\limits_{x\to2}\frac{(2-x)(\sqrt{3-x}+1)}{(2-x)(\sqrt{6-x}+2)}$ $A=\lim\limits_{x\to2}\frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$ $A=\frac{\sqrt{3-2}+1}{\sqrt{6-2}+2}$ $A=\frac{1}{2}$