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

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

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

0

Solution steps

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

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

= \frac{1}{2} \cdot x \to \infty lim (\frac{lo g _{2} ( lo g _{2} ( x ) )}{lo g _{2} ( x )})

Apply L'Hopital's Rule

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

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

Simplify

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

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

x \to \infty lim (1) = 1

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

= \frac{1}{2} \cdot \frac{1}{\infty}

Simplify

\frac{1}{2} \cdot \frac{1}{\infty} : 0

= 0

\frac{d}{d x} (2 x^{2} - 4 x + 1)

5 - 3 x

\frac{d}{d x} (\frac{6}{( 1 - x ) ^{4}})

x \to 0 lim (x - tan (x))

f (x) = \frac{x}{6 + x}

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