# Chapter 3 - Determining Change: Derivatives - 3.7 Activities - Page 244: 13

Indeterminate form is $\dfrac{0}{0}$ and $$\lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} =\frac{1}{40}$$

#### Work Step by Step

Given $$\lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25}$$ using the method of replacement \begin{align*} \lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} &=\lim _{x \rightarrow 5} \frac{(5-1)^{0.5}-2}{25-25}\\ &=\frac{0}{0} \end{align*} This is an indeterminate form, then applying L'Hôpital's Rule, we get \begin{align*} \lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} &= \lim _{x \rightarrow 5} \frac{0.5(x-1)^{-0.5}}{2x} \\ &= \lim _{x \rightarrow 5} \frac{0.5(5-1)^{-0.5}}{2(5)}\\ &= \lim _{x \rightarrow 5}\frac{1}{40}\\ &=\frac{1}{40} \end{align*}

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