Question

$\underset{x\to 4}{lim}\frac{x^3-64}{x^2-16}$

• A. $1$
• B. $0$
• C. $6$
• D. $2$

Answer 1

Answer: (C)

$6$

${lim}_{x\to 4}\frac{x^3-64}{x^2-16}$

$\therefore \frac{0}{0}$ form

$\therefore {lim}_{x\to 4}\frac{x^3-64}{x^2-16}$=${lim}_{x\to 4}\frac{\frac{d}{dx}(x^3-64)}{\frac{d}{dx}(x^2-16)}$

$\Rightarrow {lim}_{x\to 4}\frac{3x^2}{2x}$

$\Rightarrow \frac{3({lim}_{x\to 4}x{)}^2}{2{lim}_{x\to 4}x}=\frac{3(4{)}^2}{2\cdot 4}=6$