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求极限:lim(x+e的x次方)2/x次方
人气:422 ℃ 时间:2020-06-20 12:22:31
解答
lim(x→0) (x+e^x)^(2/x)
= lim(x→0) [1+(x+e^x-1)]^(2/x)
= lim(x→0) [1+(x+e^x-1)]^[1/(x+e^x-1) * (x+e^x-1)*(2/x)] 0) (1+y)^1/y=e,y=x+e^x-1
= e^2lim(x→0) (x+e^x-1)/x
= e^2lim(x→0) [1 + (e^x-1)/x] 0) (e^x-1)/x = 1
= e^[2(1+1)]
= e⁴
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