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

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Answer 1 · 2014-09-17T11:43:36

First factor the denominator...

$(x^3-64)/((x-4)(x-4))$

Now factor the numerator...

$((x-4)(x^2+4x+16))/((x-4)(x-4))$

Divide numerator and denominator by x-4...

$(x^2+4x+16)/(x-4)$

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

$((4)^2+4(4)+16)/((4)-4)$

Combine terms...

$48/0$

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

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