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

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

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

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

0

Solution steps

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

Apply L'Hopital's Rule

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

Simplify

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

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

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

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

Simplify

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

= 0

f (x) = 2 x^{4}

\int \frac{x ^{2} - 1}{x ^{2}} d x

\int x^{2} e^{- 18 x} d x

(2 x^{4} + y^{4}) d y - x^{3} y d x = 0

d er i v a t i v e f (x) = y

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