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

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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

= \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

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