# Calculus Help Stuck Part 2?

Evaluate the lim

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

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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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