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人気のある微分積分 >

limit as x approaches 0+of e^{-1/x}ln(x)

解

x \to 0 + lim (e^{- \frac{1}{x}} ln (x))

解

0

解答ステップ

x \to 0 + lim (e^{- \frac{1}{x}} ln (x))

ロピタル向けに書き換える

= x \to 0 + lim (\frac{ln ( x )}{\frac{1}{e ^{- \frac{1}{x}}}})

ロピタルの定理を適用する

= x \to 0 + lim \frac{\frac{1}{x}}{- \frac{e ^{\frac{1}{x}}}{x ^{2}}}

簡素化

\frac{\frac{1}{x}}{- \frac{e ^{\frac{1}{x}}}{x ^{2}}} : - \frac{x}{e ^{\frac{1}{x}}}

= x \to 0 + lim (- \frac{x}{e ^{\frac{1}{x}}})

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

= - x \to 0 + lim (\frac{x}{e ^{\frac{1}{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{lim _{x \to 0 +} ( x )}{lim _{x \to 0 +} ( e ^{\frac{1}{x}} )}

x \to 0 + lim (x) = 0

x \to 0 + lim (e^{\frac{1}{x}}) = \infty

= - \frac{0}{\infty}

無限大の特性を適用する:

\frac{0}{\infty} = 0

= 0

グラフ

人気の例

limit as x approaches-3+of (2-x)/(x+3)

x \to - 3 + lim (\frac{2 - x}{x + 3})

limit as x approaches 8 of 10

x \to 8 lim (10)

limit as x approaches-infinity of f(x)x

x \to - \infty lim (f (x) x)

limit as x approaches 0+of 1/(|ln(x)|)

x \to 0 + lim (\frac{1}{∣ ln ( x ) ∣})

limit as x approaches 0+of x^{(e^x-1)}

x \to 0 + lim (x^{(e^{x} - 1)})