Question

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

Answer 1

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

Applying L-Hospital's principle

i.e., $\underset{x\to 4}{lim}\frac{3x^2}{2x}$ $[i.e.,\frac{\frac{d}{dx}\left(Numerator\right)}{\frac{d}{dx}\left(denominator\right)}]$

$=\underset{x\to 4}{lim}\frac{3}{2}x$

$=\frac{3}{2}\times 4=6$.