limit as x approaches infinity of (e^{2x}-e^x)/(ln(x+1))

解

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

\infty

解答ステップ

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

ロピタルの定理を適用する

= x \to \infty lim (\frac{e ^{2 x} \cdot 2 - e ^{x}}{\frac{1}{x + 1}})

簡素化

= x \to \infty lim ((2 e^{2 x} - e^{x}) (x + 1))

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

不定形式の場合を除く

= x \to \infty lim (2 e^{2 x} - e^{x}) \cdot x \to \infty lim (x + 1)

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

x \to \infty lim (x + 1) = \infty

= \infty \cdot \infty

無限大の特性を適用する:

\infty \infty = \infty

= \infty