Answer
$-7$
Work Step by Step
We want to find $\lim\limits_{x \to -5}\dfrac{x^2+3x-10}{x+5}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches -5 is zero, and if we try to substitute $x=-5$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $x+5$ in both the numerator and the denominator.
Note that $\dfrac{x^2+3x-10}{x+5}=\dfrac{(x+5)(x-2)}{x+5}=x-2$ for all $x \neq -5.$
Thus $\lim\limits_{x \to -5}\dfrac{x^2+3x-10}{x+5}=\lim\limits_{x \to -5}x-2=-5-2=-7.$
