What is the limit as x approaches infinity of (((log_{2}(x))^2))/(sqrt(x)) ?

The limit as x approaches infinity of (((log_{2}(x))^2))/(sqrt(x)) is 0

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limit as x approaches infinity of (((log_{2}(x))^2))/(sqrt(x))

x \to \infty lim \frac{( ( lo g _{2} ( x ) ) ^{2} )}{x}

0

Solution steps

x \to \infty lim (\frac{( lo g _{2} ( x ) ) ^{2}}{x})

Refine

= x \to \infty lim (\frac{lo g _{2} ( x ) ^{2}}{x})

Apply L'Hopital's Rule

: x \to \infty lim \frac{\frac{2 l o g _{2} ( x )}{l n ( 2 ) x}}{\frac{1}{2 x}}

= x \to \infty lim \frac{\frac{2 l o g _{2} ( x )}{l n ( 2 ) x}}{\frac{1}{2 x}}

Simplify

\frac{\frac{2 l o g _{2} ( x )}{l n ( 2 ) x}}{\frac{1}{2 x}} : \frac{4 lo g _{2} ( x )}{ln ( 2 ) x}

= x \to \infty lim (\frac{4 lo g _{2} ( x )}{ln ( 2 ) x})

Apply L'Hopital's Rule

: x \to \infty lim \frac{\frac{4}{l n ( 2 ) x}}{\frac{l n ( 2 )}{2 x}}

= x \to \infty lim \frac{\frac{4}{l n ( 2 ) x}}{\frac{l n ( 2 )}{2 x}}

Simplify

\frac{\frac{4}{l n ( 2 ) x}}{\frac{l n ( 2 )}{2 x}} : \frac{8}{ln ^{2} ( 2 ) x}

= x \to \infty lim (\frac{8}{ln ^{2} ( 2 ) x})

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

= \frac{8}{ln ^{2} ( 2 )} \cdot x \to \infty lim (\frac{1}{x})

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

x \to \infty lim (1) = 1

x \to \infty lim (x) = \infty

= \frac{8}{ln ^{2} ( 2 )} \cdot \frac{1}{\infty}

Simplify

\frac{8}{ln ^{2} ( 2 )} \cdot \frac{1}{\infty} : 0

= 0

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

f (x) = x^{9} cos (x)

t^{2} e^{- 3 t}

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