What is the limit as x approaches infinity of (ln(x))/(\sqrt[3]{x)} ?

The limit as x approaches infinity of (ln(x))/(\sqrt[3]{x)} is 0

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limit as x approaches infinity of (ln(x))/(\sqrt[3]{x)}

x \to \infty lim (\frac{ln ( x )}{3 x})

0

Solution steps

x \to \infty lim (\frac{ln ( x )}{3 x})

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{1}{x}}{\frac{1}{3 x ^{\frac{2}{3}}}} : \frac{3}{x ^{\frac{1}{3}}}

= x \to \infty lim (\frac{3}{x ^{\frac{1}{3}}})

Apply Infinity Property:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 0

\frac{1}{x - 1}

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x + 1 y)

f (x) = 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 4

\frac{d y}{d x}, y^{3} + y = x

\int \frac{1}{x ^{13}} d x

What is the limit as x approaches infinity of (ln(x))/(\sqrt[3]{x)} ?
The limit as x approaches infinity of (ln(x))/(\sqrt[3]{x)} is 0