Question

\(\lim\limits_{x\rightarrow-3}\dfrac{x^3+27}{2x^2+3x-9}\)

Answer 1

\(\lim\limits_{x\rightarrow-3}\dfrac{x^3+27}{2x^2+3x-9}=\lim\limits_{x\rightarrow-3}\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{\left(x+3\right)\left(2x-3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x+9}{2x-3}=-3\)