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The limit expression [tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h}[/tex] is 1/12
The limit expression is given as [tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h}[/tex]
Apply the L'hopital rule (i.e. differentiate the numerator and the denominator, separately)
The derivative of the numerator is [tex]\frac{1}{3 * \left(h + 8\right)^{\frac{2}{3}}}[/tex] and the denominator is 1
So, the expression can be rewritten as follows
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \lim_{h \to 0} \dfrac{\frac{1}{3 * \left(h + 8\right)^{\frac{2}{3}}}}{1}[/tex]
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \lim_{h \to 0} \frac{1}{3 * \left(h + 8\right)^{\frac{2}{3}}}[/tex]
Substitute 0 for h
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \dfrac{1}{3 \left(0 + 8\right)^{\frac{2}{3}}}[/tex]
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \dfrac{1}{3 * \left(8\right)^{\frac{2}{3}}}[/tex]
Take the cube root of 8
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \dfrac{1}{3 * \left(2\right)^{2}}[/tex]
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \dfrac{1}{3*4}[/tex]
[tex]\lim_{h \to 0} \dfrac{(8 + h)^\frac13 - 2}{h} = \dfrac{1}{12}[/tex]
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Rewritten by : Barada
