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.

Cube root of a real number $p$ is the unique real number $q$ such that $q^3=p$. Therefore, $x^{\frac{1}{3}} \to -1$ as $x \to -1$ because $(-1)^3=-1$ and so $(-1)^{\frac{1}{3}}=-1$.
