## Thomas' Calculus 13th Edition

$-10$
Consider: $\lim\limits_{x \to -5}f(x)=\lim\limits_{x \to -5}\dfrac{x^2-25}{x +5}$ Need to check that the limit has an indeterminate form. Thus, $f(-5)=\dfrac{(-5)^2 -25}{-5+5}=\dfrac{0}{0}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to -5}\dfrac{2x}{1}=2(-5)=-10$