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  4. Let displaystyle lim h arrow 0((1/h √[3]8+h)-(1/2 h))=k, then find 96 k+3.

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

Limits and Derivatives

Solution:

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