limit as x approaches infinity of-xe^{1-x}

解答

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

0

求解步骤

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

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

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

化简

e^{1 - x} : e e^{- x}

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

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

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

改写以应用洛必达法则

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

使用洛必达法则

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

不定型除外

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

x \to \infty lim (1) = 1

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

= - e \frac{1}{\infty}

化简

- e \frac{1}{\infty} : 0

= 0

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