## Thomas' Calculus 13th Edition

$a.\quad +\infty$ $b.\quad -\infty$
$a.$ 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 $\displaystyle \frac{3}{t^{3/5}}\rightarrow-\infty$ $(\displaystyle \frac{3}{t^{3/5}}+7)\rightarrow-\infty$