## Calculus (3rd Edition)

$$\lim _{x \rightarrow 5} \frac{2x^2-9x-5}{x^2-25} =\frac{11}{10}$$
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}