Question

Find the following limit.
$\underset{x\to 0}{lim}\frac{(4+x{)}^3-64}{x}.$

Answer 1

$\underset{x\to 0}{lim}\frac{(x+4{)}^3-64}{x}$

$=\underset{x\to 0}{lim}\frac{(x+4{)}^3-(4{)}^3}{x}[a^3-b^3=(a-b\left)\right(a^2+b^2+ab\left)\right]$

$=\underset{x\to 0}{lim}\frac{(x-4-4)\left[\right(x+4{)}^2+4(x+4)+16]}{x}$

$=\underset{x\to 0}{lim}(x+4{)}^2+4(x+4)+16$

$=16+16+16$

$=48$.