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受欢迎的微积分 >

limit as x approaches infinity of x^5e^{-2x}

解答

x \to \infty lim (x^{5} e^{- 2 x})

解答

0

求解步骤

x \to \infty lim (x^{5} e^{- 2 x})

改写以应用洛必达法则

= x \to \infty lim (\frac{x ^{5}}{\frac{1}{e ^{- 2 x}}})

使用洛必达法则

= x \to \infty lim (\frac{5 x ^{4}}{e ^{2 x} \cdot 2})

使用洛必达法则

= x \to \infty lim (\frac{20 x ^{3}}{4 e ^{2 x}})

使用洛必达法则

= x \to \infty lim (\frac{60 x ^{2}}{8 e ^{2 x}})

使用洛必达法则

= x \to \infty lim (\frac{120 x}{16 e ^{2 x}})

使用洛必达法则

= x \to \infty lim (\frac{120}{32 e ^{2 x}})

约分:

8

= x \to \infty lim (\frac{15}{4 e ^{2 x}})

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

= \frac{15}{4} \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

不定型除外

= \frac{15}{4} \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

= \frac{15}{4} \cdot \frac{1}{\infty}

化简

\frac{15}{4} \cdot \frac{1}{\infty} : 0

= 0

作图

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