Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 60

Find $\displaystyle \lim \limits_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$



$\begin{equation} \begin{aligned} & \lim \limits_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \frac{\sqrt{6 - x} + 2}{\sqrt{6 - x} + 2} = \lim \limits_{x \to 2} \frac{6 - x - 4}{(\sqrt{3-x} - 1)(\sqrt{6 -x } + 2)} && \text{ Multiply both numerator and denominator by $\sqrt{6 - x} + 2$ and simplify.}\\ \\ & \lim \limits_{x \to 2} \frac{2 - x}{(\sqrt{3 - x} - 1)(\sqrt{6 - x + 2})} \cdot \frac{\sqrt{3 - x} + 1}{\sqrt{3 - x } + 1} = \lim \limits_{x \to 2} \frac{(2 - x)(\sqrt{3 - x} + 1)}{(3 - x - 1)(\sqrt{6 - x} + 2)} && \text{ Multiply both numerator and denominator by $\sqrt{3 - x} + 1$ and simplify.}\\ \\ & \lim \limits_{x \to 2} \frac{\sqrt{3 - x} + 1}{\sqrt{6 - x} + 2} = \frac{\sqrt{3 - 2} + 1}{\sqrt{6 - 2} + 2} \quad = \frac{\sqrt{1}+1}{\sqrt{4}+2} \quad = \frac{2}{4} \quad =\frac{1}{2} && \text{ Substitute the value of $x$ and simplify}\\ \\ &\fbox{$\lim \limits_{x \to 2} \displaystyle \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} = \frac{1}{2}$} \end{aligned} \end{equation}$