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e^(-1/x^2)/x x趋向于0 求极限
人气:323 ℃ 时间:2020-01-25 23:28:21
解答
设y = 1/x²,x = ±y^(-1/2)
e^(-1/x^2)/x
= ±e^(-y) / y^(-1/2)
= ±y^(1/2) / e^y
x → 0 等价于 y → ∞
lim[(e^(-1/x^2))/x,x → 0]
= lim[ ±y^(1/2) / e^y,y → ∞ ]
y^(1/2) / e^y 为 ∞/∞ 型,可用洛必达法则
y^(1/2)求导为(1/2)y^(-1/2),e^y求导为e^y
lim[(e^(-1/x^2))/x,x → 0]
= lim[ ±y^(1/2) / e^y,y → ∞ ]
= lim[ ±(1/2)y^(-1/2) / e^y,y → ∞ ]
= lim[ ±1 / 2y^(1/2)e^y,y → ∞ ]
= 0
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