Question

Let $\underset{h\to 0}{\lim }\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)=k$, then find $96k+3$.

Answer 1

Answer: 1

$\underset{h\to 0}{\lim }\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)=\underset{h\to 0}{\lim }\left(\frac{2-\sqrt[3]{8+h}}{2h}\sqrt[3]{8+h}\right)$
$k=\underset{h\to 0}{\lim }\left(\frac{(2-\sqrt[3]{8+h})\left(2^2+(8+h{)}^{\frac{2}{3}}+2\sqrt[3]{8+h}\right)}{2h\sqrt[3]{8+h}\left(2^2+(8+h{)}^{\frac{2}{3}}+2\sqrt[3]{8+h}\right)}\right)$
$=\underset{h\to 0}{\lim }\left(\frac{8-(8+h)}{2h\sqrt[3]{8+h}\left(4+(8+h{)}^{\frac{2}{3}}+2\sqrt[3]{8+h}\right)}\right)$
$=-\frac{1}{2}\times \frac{1}{2}\times \frac{1}{4+4+2(2)}$
$=-\frac{1}{48}$
$\Rightarrow 96k+3=-2+3=1$