Answer
$$\lim_{x\to0}\frac{1}{x^{2/3}}=\infty$$
Work Step by Step
$$A=\lim_{x\to0}\frac{1}{x^{2/3}}=\lim_{x\to0}\frac{1}{\sqrt[3]{x^2}}$$
As $x\to0$, $\sqrt[3]{x^2}$ approaches $0$ as well.
Since $x^2\gt0$ for all $x$, $\sqrt[3]{x^2}\gt0$ also for all $x$, so $1/\sqrt[3]{x^{2}}$ will approach $\infty$. Therefore,
$$A=\infty$$