Question

Evaluate $\underset{x\to 2}{lim}\frac{x^6-24x-16}{x^3+2x-12}$

• A. $10$
• B. $12$
• C. $14$
• D. $16$

Answer 1

Answer: (B)

$12$

Method I
$\underset{x\to 2}{lim}\frac{x^6-24x-16}{x^3+2x-12}$ $\left(\frac{0}{0}form\right)$
$=\underset{x\to 2}{lim}\frac{(x-2)(x^5+2x^4+4x^3+8x^2+16x+8)}{(x-2(x^2+2x+6)}$
$=\underset{x\to 2}{lim}\frac{(x^5+2x^4+4x^3+8x^2+16x+8)}{(x^2+2x+6)}$
$=\frac{2^5+2{\left(2\right)}^4+4{\left(2\right)}^3+8{\left(2\right)}^2+16\left(2\right)+8)}{{\left(2\right)}^2+2\left(2\right)+6)}$
$=\frac{168}{14}=12$
Method II (Applying L'Hospital's rule)
If $x=2$, $\frac{x^6-24x-16}{x^3+2x-12}$
$\underset{x\to 2}{lim}\frac{x^6-24x-16}{x^3+2x-12}$ $\left(\frac{0}{0}form\right)$
$=\underset{x\to 2}{lim}\frac{6x^5-24}{3x^2+2}$ (Applying L'Hospital's rule)
$=\frac{6{\left(2\right)}^5-24}{3{\left(2\right)}^2+2}$
$=\frac{168}{14}=12$