Answer
$$\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25} =\frac{11}{10}$$
Work Step by Step
Given $$\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25}$$
let $$ f(x) = \frac{2x^2-9x-5}{x^2-25}$$
Since, we have
$$ f(5)=\frac{50-45-5}{25-25}=\frac{0}{0}$$
So, transform algebraically and cancel
\begin{aligned}
L&=\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25}\\
&= \lim _{x \rightarrow 5} \frac{(x -5)(2x+1)}{(x -5)(x+5)}\\
&= \lim _{x \rightarrow 5} \frac{ (2x+1)}{ (x+5)}\\
&= \frac{10+1}{5+5}\\
&=\frac{11}{10}
\end{aligned}
