Calculus Help Stuck Part 2?

Evaluate the lim

a. lim x--> 64 (cube root x-4/x-64)

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asked by Marc
  1. Sorry meant for title to be part 3 not part 2

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    posted by Marc
  2. (∛x-4)/(x-64) -> 0/0
    so try L'Hospital's rule. Then the limit is

    (1/(3∛x^2) / (1) = 1/(3*16) = 1/48

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    posted by Steve
  3. or, consider x-64 as the sum of cubes:

    x-64 = (∛x)^3 - 4^3

    so it factors into

    (∛x-4)(∛x^2+4∛x+16)

    Now the fraction is just

    1/(∛x^2+4∛x+16) = 1/(16+16+16) = 1/48

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    posted by Steve
  4. Shouldn't i let cube root of x equal u and then factor it out like this

    u-4/u^3-64

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    posted by Marc

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