Question

Find the limit of the function algebraically. limit as x approaches negative three of quantity x squared minus nine divided by quantity x cubed plus three.

Answer 1

The limand is continuous at $x=-3$, so you can directly substitute $x=-3$ to get

$\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\frac{(-3)^2-9}{(-3)^3+3}=\dfrac{9-9}{-27+3}=0$

Did you mean to write $x^3+3^3=x^3+27$ in the denominator by any chance? In that case, you would instead have

$\displaystyle\lim_{x\to-3}\frac{x^2-9}{x^3+3}=\lim_{x\to-3}\frac{(x+3)(x-3)}{(x+3)(x^2-3x+9)}=\lim_{x\to-3}\frac{x-3}{x^2-3x+9}=\frac{-3-3}{(-3)^2-3(-3)+9}=-\frac29$