Question

Solve $\underset{x\to 3}{lim}\frac{x^3+x^2-12x}{x^2-9}$.

• A. $-7$
• B. $-\frac{7}{2}$
• C. $0$
• D. $\frac{7}{2}$
• E. The limit does not exist

Answer 1

The correct option is D $\frac{7}{2}$

Use L-hospital, we get

$\underset{x\to 3}{lim}\frac{3x^2+2x-12}{2x-0}$

$=\underset{x\to 3}{lim}\frac{3(3{)}^2+2(3)-12}{2\left(3\right)-0}=\frac{21}{6}=\frac{7}{2}$