# Find B such that $\lim_{x\rightarrow-3^-}\frac{x^2-4}{x^2+7x+B}=-\infty$

Okay so I know that the answer is 12 but I don't understand how to get there. Am I right in thinking that we're going to use l hospitals rule for this? Can someone please guide me and explain to me the process of solving this problem? Thank you!

First note that $$\lim_{x\rightarrow-3}x^2-4=5>0$$ and in this case it doesn't matter if the limit approaches from right or left.
In order to get infinity in the limit above, $-3$ must be a zero of $$x^2+7x+B$$ So we have to impose $(-3)^2+7(-3)+B=0$ from which we get $B=12$.