limit as x approaches infinity of-2xe^{-2x}

解

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

0

解答ステップ

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

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

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

ロピタル向けに書き換える

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

ロピタルの定理を適用する

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

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

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

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

x \to \infty lim (1) = 1

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

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

簡素化

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

= 0