## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{t\to0^+}\Big(\frac{1}{t^{3/5}}+7\Big)=\infty$$ $$\lim_{t\to0^-}\Big(\frac{1}{t^{3/5}}+7\Big)=-\infty$$
(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, $$B=-\infty$$