Answer
$$\lim_{t\to0^+}\Big(2-\frac{3}{t^{1/3}}\Big)=-\infty$$
$$\lim_{t\to0^+}\Big(2-\frac{3}{t^{1/3}}\Big)=\infty$$
Work Step by Step
(a) $$A=\lim_{t\to0^+}\Big(2-\frac{3}{t^{1/3}}\Big)=\lim_{t\to0^+}\frac{2t^{1/3}-3}{t^{1/3}}$$
As $t\to0^+$:
$t^{1/3}\to0^+\gt0$
$2t^{1/3}-3\to-3^+\lt0$
So $\frac{2t^{1/3}-3}{t^{1/3}}\lt0$, and the function will approach $-\infty$. Therefore, $$A=-\infty$$
(b) $$B=\lim_{t\to0^-}\Big(2-\frac{3}{t^{1/3}}\Big)=\lim_{t\to0^-}\frac{2t^{1/3}-3}{t^{1/3}}$$
As $t\to0^-$:
$t^{1/3}\to0^-\lt0$
$2t^{1/3}-3\to-3^-\lt0$
So $\frac{2t^{1/3}-3}{t^{1/3}}\gt0$, and the function will approach $\infty$. Therefore, $$B=\infty$$
