Answer
Indeterminate form is $\dfrac{0}{0}$ and
$$\lim _{x \rightarrow 7} \frac{x^{2}-2 x-35}{7 x-x^{2}} =\frac{-12 }{7 }$$
Work Step by Step
Given $$\lim _{x \rightarrow 7} \frac{x^{2}-2 x-35}{7 x-x^{2}}$$
using the method of replacement
\begin{align*}
\lim _{x \rightarrow 7} \frac{x^{2}-2 x-35}{7 x-x^{2}}&=\lim _{x \rightarrow 7} \frac{49-14-35}{49-49}\\
&=\lim _{x \rightarrow 7} \frac{0}{0}
\end{align*}
This is an indeterminate form, then applying L'Hôpital's Rule, we get
\begin{align*}
\lim _{x \rightarrow 7} \frac{x^{2}-2 x-35}{7 x-x^{2}}&=\lim _{x \rightarrow 7} \frac{2x-2 }{7 -2x}\\
&=\lim _{x \rightarrow 7} \frac{14-2 }{7 -14}\\
&=\frac{-12 }{7 }
\end{align*}
