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

The limit as x approaches infinity of (log_{2}(x))/(\sqrt[3]{x)} is 0

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limit as x approaches infinity of (log_{2}(x))/(\sqrt[3]{x)}

0

Solution steps

Apply L'Hopital's Rule

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

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

Simplify

\frac{\frac{1}{x l n ( 2 )}}{\frac{1}{3 x ^{\frac{2}{3}}}} : \frac{3}{ln ( 2 ) x ^{\frac{1}{3}}}

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

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

x \to \infty lim (1) = 1

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

= \frac{3}{ln ( 2 )} \cdot \frac{1}{\infty}

Simplify

\frac{3}{ln ( 2 )} \cdot \frac{1}{\infty} : 0

= 0

x \to 0 lim (2 + \frac{5}{x ^{2}})

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

cos (x), sin (x), [0, \frac{π}{2}]

\int x^{5} 9 x^{2} - 1 d x

\frac{\partial}{\partial y} (3 x^{2} y + cos (y))

What is the limit as x approaches infinity of (log_{2}(x))/(\sqrt[3]{x)} ?
The limit as x approaches infinity of (log_{2}(x))/(\sqrt[3]{x)} is 0