Evaluate the lim
a. lim x--> 64 (cube root x-4/x-64)
4 answers
Sorry meant for title to be part 3 not part 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
so try L'Hospital's rule. Then the limit is
(1/(3∛x^2) / (1) = 1/(3*16) = 1/48
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
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
Shouldn't i let cube root of x equal u and then factor it out like this
u-4/u^3-64
u-4/u^3-64
1 answer