# Chapter 2 - Limits - 2.5 Evaluating Limits Algebraically - Exercises - Page 72: 23

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

#### Work Step by Step

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

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