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Answer :

[tex]\begin{gathered} \lim _{h\to0}\frac{\sqrt[3]{8+h}-2}{h} \\ \text{Choosing number near h from left} \\ h=-1 \\ \frac{\sqrt[3]{8-1}-2}{-1}=0.087 \\ h=-0.5 \\ \frac{\sqrt[3]{8-0.5}-2}{-0.5}=0.0851 \\ h=-0.005 \\ \frac{\sqrt[3]{8-0.005}-2}{-0.005}=0.08335 \\ h=-0.0005 \\ \frac{\sqrt[3]{8-0.0005}-2}{-0.0005}=0.08333 \\ \\ \text{Choosing number near h from right} \\ h=1 \\ \frac{\sqrt[3]{8+1}-2}{1}=0.08 \\ h=0.5 \\ \frac{\sqrt[3]{8+0.5}-2}{0.5}=0.0816 \\ h=0.005 \\ \frac{\sqrt[3]{8+0.005}-2}{0.005}=0.0833 \\ h=0.0005 \\ \frac{\sqrt[3]{8+0.0005}-2}{0.0005}=0.08333 \\ \text{From the values found},\text{ it would say} \\ \lim _{h\to0}\frac{\sqrt[3]{8+h}-2}{h}=0.08333\approx\frac{1}{12} \end{gathered}[/tex]