Question

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

Answer 1

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

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

$={lim}_{x\to 4}\frac{\frac{x^3-4^3}{x-4}}{\frac{x^2-4^2}{x-4}}$ [Dividing numerator and denomiantor by x-4]

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

Applying ${lim}_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$ in numerator and denominator

$\Rightarrow \frac{{lim}_{x\to 4}\frac{x^3-4^3}{x-4}}{{lim}_{x\to 4}\frac{x^2-4^2}{x-4}}=\frac{3(4{)}^{3-1}}{2(4{)}^{2-1}}=\frac{3(4{)}^2}{2\left(4\right)}$

=6