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Question
\[\lim_{h \to 0} \left\{ \frac{1}{h\sqrt[3]{8 + h}} - \frac{1}{2h} \right\} =\]
Options
−1/12
−4/3
−16/3
−1/48
Solution
−1/48
\[\lim_{h \to 0} \left[ \frac{1}{h \sqrt[3]{8 + h}} - \frac{1}{2h} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{1}{\sqrt[3]{8 + h}} - \frac{1}{2} \right\} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{2 - \left( 8 + h \right)^{1/3}}{2 \times \sqrt[3]{8 + h}} \right\} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{1}{h}\left\{ \frac{8^{1/3} - \left( 8 + h \right)^{1/3}}{2 \sqrt[3]{8 + h}} \right\} \right] \left[ A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right) or A - B = \frac{A^3 - B^3}{A^2 + AB + B^2} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{8 - \left( 8 + h \right)}{h\left\{ 2\sqrt[3]{8 + h} \right\}\left\{ 4 + 2 \left( 8 + h \right)^{1/3} + \left( 8 + h \right)^{2/3} \right\}} \right]\]
\[ = \left[ \frac{- 1}{2 \times \sqrt[3]{8}\left\{ 4 + 2 \times 8^{1/3} + 8^{2/3} \right\}} \right]\]
\[ = \frac{- 1}{2 \times 2\left( 4 + 4 + 4 \right)}\]
\[ = \frac{- 1}{48}\]
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