Answer
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*}