Evaluate the lim
a. lim x--> 64 (cube root x-4/x-64)
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
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
Shouldn't i let cube root of x equal u and then factor it out like this
u-4/u^3-64
To evaluate the limit of the function as x approaches 64, we will first simplify the function and then substitute the value of x.
The given function is (cube root x - 4) / (x - 64).
To simplify this expression, we can use the identity:
a^3 - b^3 = (a - b)(a^2 + ab + b^2).
In our case, a = cube root x and b = 4. Therefore, we have:
(cube root x)^3 - 4^3 = (cube root x - 4)(cube root x^2 + 4(cube root x) + 16).
Now, let's substitute this expression into our fraction:
(cube root x - 4) / (x - 64) = [(cube root x)^3 - 4^3] / (x - 64)
= [(cube root x - 4)(cube root x^2 + 4(cube root x) + 16)] / (x - 64).
Now, we can cancel out common factors between the numerator and denominator. Since (cube root x - 4) is a common factor, we are left with:
(cube root x^2 + 4(cube root x) + 16) / (x - 64).
To evaluate the limit, we substitute x = 64 into the simplified expression:
(cube root 64^2 + 4(cube root 64) + 16) / (64 - 64).
Simplifying further, we have:
(8^2 + 4(8) + 16) / 0 = (64 + 32 + 16) / 0.
However, this expression is undefined because we cannot divide by zero. Therefore, the limit does not exist.