Question

$\underset{h\to 0}{\lim }\left\{\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right\}=$

(a) −1/12
(b) −4/3
(c) −16/3
(d) −1/48

Answer 1

(d) −1/48

$$\begin{gathered} \underset{h\to 0}{\lim }\left[\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{1}{h}\left\{\frac{1}{\sqrt[3]{8+h}}-\frac{1}{2}\right\}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{1}{h}\left\{\frac{2-{\left(8+h\right)}^{1/3}}{2\times \sqrt[3]{8+h}}\right\}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{1}{h}\left\{\frac{8^{1/3}-{\left(8+h\right)}^{1/3}}{2\sqrt[3]{8+h}}\right\}\right]\left[A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)orA-B=\frac{A^3-B^3}{A^2+AB+B^2}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{8-\left(8+h\right)}{h\left\{2\sqrt[3]{8+h}\right\}\left\{4+2{\left(8+h\right)}^{1/3}+{\left(8+h\right)}^{2/3}\right\}}\right] \\ =\left[\frac{-1}{2\times \sqrt[3]{8}\left\{4+2\times 8^{1/3}+8^{2/3}\right\}}\right] \\ =\frac{-1}{2\times 2\left(4+4+4\right)} \\ =\frac{-1}{48} \end{gathered}$$