# Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 22

$$\lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8} =\frac{1}{4}$$

#### Work Step by Step

Given $$\lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8}$$ let $$f(x) = \lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8}$$ Since, we have $$f(8)= \frac{\sqrt{8-4}-2}{8-8}=\frac{0}{0}$$ So, transform algebraically and cancel \begin{aligned}L&= \lim _{x \rightarrow 8} \frac{\sqrt{x-4}-2}{x-8} \\ &=\lim _{x \rightarrow 4} \frac{\sqrt{x-4}-2}{x-8} \times \frac{\sqrt{x-4}+2}{\sqrt{x-4}+2}\\ &=\lim _{x \rightarrow 4} \frac{(\sqrt{x-4})^{2}-2^{2}}{(x-8)(\sqrt{x-4}+2)}\\ &= \lim _{x \rightarrow 4} \frac{(x-4-4)}{(x-8)(\sqrt{x-4}+2)}\\ &=\lim _{x \rightarrow 4} \frac{(x-8)}{(x-8)(\sqrt{x-4}+2)} \\ &=\lim _{x \rightarrow 4} \frac{1}{\sqrt{x-4}+2} \\ &=\frac{1}{\sqrt{8-4}+2} \\ &=\frac{1}{\sqrt{4}+2} \\ &=\frac{1}{2+2}\\ &=\frac{1}{4} \end{aligned}

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