What is the limit as x approaches infinity of (2x)/(2^xln(2)) ?

The limit as x approaches infinity of (2x)/(2^xln(2)) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches infinity of (2x)/(2^xln(2))

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

0

Solution steps

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

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{x}{2 ^{x}})

Apply L'Hopital's Rule

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

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

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

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

x \to \infty lim (1) = 1

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

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

Simplify

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

= 0

x \to 1 - lim (\frac{x - 2 x ^{- 1}}{4 x ^{- 1}})

x \to 0 lim (x \cdot ln^{3} (x))

x \to 0 lim (∣ x - 2 ∣)

x \to 3 lim (x^{2} + 9 x - 27)

x \to \infty lim (\frac{v ^{2} + 1}{3 v ^{3} - 3})

What is the limit as x approaches infinity of (2x)/(2^xln(2)) ?
The limit as x approaches infinity of (2x)/(2^xln(2)) is 0