Question

Evaluate: $\underset{x\to -3}{lim}\frac{x^3+27}{x+3}$

• A. $0$
• B. $9$
• C. $27$
• D. $-3$
• E. Undefined

Answer 1

Answer: (C)

$27$

$\underset{x\to -3}{lim}\frac{x^3+3^3}{x+3}=\underset{x\to -3}{lim}\frac{(x+3)(x^2+3^2-3x)}{(x+3)}$

$[a^3+b^3=(a+b)(a^2+b^2-ab)]$

$=\underset{x\to -3}{lim}(x^2+9-3x)={(-3)}^2+9-3(-3)$

$=27$