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.