What is the limit as x approaches-8 of (\sqrt[3]{x}+2)/(x+8) ?

The limit as x approaches-8 of (\sqrt[3]{x}+2)/(x+8) is 1/(12(-1)^{2/3)}

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limit as x approaches-8 of (\sqrt[3]{x}+2)/(x+8)

\frac{1}{12 ( - 1 ) ^{\frac{2}{3}}}

Solution steps

Apply L'Hopital's Rule

: x \to - 8 lim (\frac{\frac{1}{3 x ^{\frac{2}{3}}}}{1})

= x \to - 8 lim (\frac{\frac{1}{3 x ^{\frac{2}{3}}}}{1})

Plug in the value

x = - 8

= \frac{\frac{1}{3 ( - 8 ) ^{\frac{2}{3}}}}{1}

Simplify

\frac{\frac{1}{3 ( - 8 ) ^{\frac{2}{3}}}}{1} : \frac{1}{12 ( - 1 ) ^{\frac{2}{3}}}

= \frac{1}{12 ( - 1 ) ^{\frac{2}{3}}}

\int \frac{x ^{2} + 3 x + 7}{x} d x

y = 5 x^{2} e^{2 x}

y = 4 \cdot 5^{x}

\frac{d}{d x} (\frac{x ( x ^{2} - 1 )}{x ^{2} + 1})

x \to 0 lim (x^{2} (sin (\frac{1}{x})))

What is the limit as x approaches-8 of (\sqrt[3]{x}+2)/(x+8) ?
The limit as x approaches-8 of (\sqrt[3]{x}+2)/(x+8) is 1/(12(-1)^{2/3)}