## Calculus (3rd Edition)

$$\lim _{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}}=2$$
Given $$\lim _{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$ let $$f(x) = \frac{\sqrt{5-x}-1}{2-\sqrt{x}}$$ Since, we have $$f(4)=\frac{1-1}{2-2}=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned} L&= \lim _{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}} \\ &= \lim _{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}} \times \frac{\sqrt{5-x}+1}{\sqrt{5-x}+1}\\ &= \lim _{x \rightarrow 4} \frac{(\sqrt{5-x})^{2}-1^{2}}{(2-\sqrt{x})(\sqrt{5-x}+1)}\\ &= \lim _{x \rightarrow 4} \frac{5-x-1}{(2-\sqrt{x})(\sqrt{5-x}+1)}\\ &= \lim _{x \rightarrow 4} \frac{4-x}{(2-\sqrt{x})(\sqrt{5-x}+1)}\\ &= \lim _{x \rightarrow 4} \frac{2^{2}-(\sqrt{x})^{2}}{(2-\sqrt{x})(\sqrt{5-x}+1)}\\ &= \lim _{x \rightarrow 4} \frac{(2+\sqrt{x})(2-\sqrt{x})}{(2-\sqrt{x})(\sqrt{5-x}+1)}\\ &= \lim _{x \rightarrow 4} \frac{2+\sqrt{x}}{\sqrt{5-x}+1}\\ &= \frac{2+\sqrt{4}}{\sqrt{5-4}+1}\\ &= \frac{2+2}{\sqrt{1}+1}\\ &= \frac{4}{2}\\ &=2 \end{aligned}