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

Question

Samantha Adkison · Accepted Answer

#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#

Adam Adkins · Answer

undefined To find the limit of (x^2 - 64) / ((x^(1/3)) -2) as x approaches 8, we can substitute 8 into the expression and simplify. Plugging in 8, we get (8^2 - 64) / ((8^(1/3)) -2). Simplifying further, we have (64 - 64) / (2 - 2), which results in 0/0. This is an indeterminate form. To evaluate the limit, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (2x) / (1/3 * x^(-2/3)). Simplifying this, we have 6x^(5/3). Plugging in 8, we get 6 * 8^(5/3), which equals 48. Therefore, the limit of (x^2 - 64) / ((x^(1/3)) -2) as x approaches 8 is 48.