Answer
$$\lim_{t\to0^+}\Big(\frac{1}{t^{3/5}}+7\Big)=\infty$$
$$\lim_{t\to0^-}\Big(\frac{1}{t^{3/5}}+7\Big)=-\infty$$
Work Step by Step
(a) $$A=\lim_{t\to0^+}\Big(\frac{1}{t^{3/5}}+7\Big)=\lim_{t\to0^+}\frac{1+7t^{3/5}}{t^{3/5}}$$
As $t\to0^+$:
$t^{3/5}\to0^+\gt0$
$(1+7t^{3/5})\to1^+\gt0$
So $\frac{1+7t^{3/5}}{t^{3/5}}\gt0$, and the function will approach $\infty$. Therefore, $$A=\infty$$
(b) $$B=\lim_{t\to0^-}\Big(\frac{1}{t^{3/5}}+7\Big)=\lim_{t\to0^-}\frac{1+7t^{3/5}}{t^{3/5}}$$
As $t\to0^-$:
$t^{3/5}\to0^-\lt0$
$(1+7t^{3/5})\to1^-\gt0$
So $\frac{1+7t^{3/5}}{t^{3/5}}\lt0$, and the function will approach $-\infty$. Therefore, $$B=-\infty$$
