limit as n approaches infinity of 5ne^{-n}

解

n \to \infty lim (5 n e^{- n})

0

解答ステップ

n \to \infty lim (5 n e^{- n})

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

= 5 \cdot n \to \infty lim (n e^{- n})

ロピタル向けに書き換える

= 5 \cdot n \to \infty lim (\frac{n}{\frac{1}{e ^{- n}}})

ロピタルの定理を適用する

= 5 \cdot n \to \infty lim (\frac{1}{e ^{n}})

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

= 5 \cdot \frac{lim _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( e ^{n} )}

n \to \infty lim (1) = 1

n \to \infty lim (e^{n}) = \infty

= 5 \cdot \frac{1}{\infty}

簡素化

5 \cdot \frac{1}{\infty} : 0

= 0