Evaluating an improper integral - issues taking the cubic root of a negative number

Problem:
Evaluate the following integral.
$$ int_{-1}^{-1} frac{dx}{x^frac{2}{3}} $$

Answer:

This integral includes the point $x = 0$ which results in a division by $0$. To get around this difficulty, we break the integral into two integrals.
begin{align*}
int_{-1}^{-1} frac{dx}{x^frac{2}{3}} &= int_{-1}^{0} frac{dx}{x^frac{2}{3}} + int_{0}^{1} frac{dx}{x^frac{2}{3}} \
int_{-1}^{0} frac{dx}{x^frac{2}{3}} &= int_{-1}^{0} x^{-frac{2}{3}} ,,, dx \
int_{-1}^{0} frac{dx}{x^frac{2}{3}} &= 3x^{frac{1}{3}} Big|_{-1}^0
= lim_{x to 0} 3x^{frac{1}{3}} – lim_{x to -1} 3x^{frac{1}{3}} \
lim_{x to 0} 3x^{frac{1}{3}} &= 0 \
end{align*}
Am I right so far? I do not know how to evaluate the following limit:
$$ lim_{x to -1} 3x^{frac{1}{3}} $$
The problem is taking the cube root of a negative number.