Answer
$ a.\quad +\infty$
$ b.\quad -\infty$
Work Step by Step
$a.$
When $t\rightarrow 0^{+}$, the denominator in $\displaystyle \frac{3}{t^{3/5}}$ is positive, approaching zero, so $\displaystyle \frac{3}{t^{3/5}}\rightarrow+\infty$
$(\displaystyle \frac{3}{t^{3/5}}+7)\rightarrow+\infty$
$b.$
When $t\rightarrow 0^{-}$, the denominator in $\displaystyle \frac{3}{t^{3/5}}$ is negative, approaching zero, so $\displaystyle \frac{3}{t^{3/5}}\rightarrow-\infty$
$(\displaystyle \frac{3}{t^{3/5}}+7)\rightarrow-\infty$