∵ cos(8π/x) ≤ 1 (x≠0) ,e^x 单调递增;
∴ 0 < e^(cos(8π/x)) ≤ e^1 = e
∴ 0 < [(3x)^1/2] * e^(cos(8π/x) ≤ [(3x)^1/2]
∵ lim(x->0+) [(3x)^1/2] = 0 ,由夹逼定理:
∴ lim(x->0+) [(3x)^1/2] * e^(cos(8π/x) = 0
Lim[(3x)^1/2] * e^(cos(8pi/ x) ( x-->+0) 求极限?
Lim[(3x)^1/2] * e^(cos(8pi/ x) ( x-->+0) 求极限?
数学人气:400 ℃时间:2020-02-03 16:14:13
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