How do you find the limit of #(x^2 - 64) / ((x^(1/3)) -2)# as x approaches #8#?

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Answer 1 · 2016-09-11T22:27:28

#192#

First, factor the numerator as a difference of squares, which takes the form #a^2-b^2=(a+b)(a-b)#.

#lim_(xrarr8)(x^2-8^2)/(x^(1/3)-2)#

#=lim_(xrarr8)((x+8)(x-8))/(x^(1/3)-2)#

Now, factor #x-8# as a difference of cubes, which takes the form #a^3-b^3=(a-b)(a^2+ab+b^2)#.

#=lim_(xrarr8)((x+8)((x^(1/3))^3-2^3))/(x^(1/3)-2)#

#=lim_(xrarr8)((x+8)(x^(1/3)-2)(x^(2/3)+2x^(1/3)+4))/(x^(1/3)-2)#

The #x^(1/3)-2# terms will cancel:

#=lim_(xrarr8)((x+8)(x^(2/3)+2x^(1/3)+4))#

Plug in #8#:

#=(8+8)(8^(2/3)+2(8^(1/3))+4)#

#=16(4+2(2)+4)#

#=16(12)#

#=192#