How do you find the limit lim_(x->4)(x^3-64)/(x^2-8x+16) ?

Sep 17, 2014

First factor the denominator...

$\frac{{x}^{3} - 64}{\left(x - 4\right) \left(x - 4\right)}$

Now factor the numerator...

$\frac{\left(x - 4\right) \left({x}^{2} + 4 x + 16\right)}{\left(x - 4\right) \left(x - 4\right)}$

Divide numerator and denominator by x-4...

$\frac{{x}^{2} + 4 x + 16}{x - 4}$

Replace all x's with the limit being approached (4)...

$\frac{{\left(4\right)}^{2} + 4 \left(4\right) + 16}{\left(4\right) - 4}$

Combine terms...

$\frac{48}{0}$

The limit approaches infinity since division by 0 is undefined, but division by 0 also approaches infinity.