limit as x approaches infinity of (2x-1)/(2^x)

解

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

0

解答ステップ

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

ロピタルの定理を適用する

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

不定形式の場合を除く

= \frac{2}{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}{\infty}

簡素化

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

= 0