這篇不是要問問題啦XD
是我打算把數學板一些不錯的題目慢慢地在這建檔0.0
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Show that lim xlnx = 0 without L'hospital Rule
x->0+
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Let x = e^(-t)
lim xlnx = lim (-t)e^(-t) = - lim t/e^t
x->0+ t->∞ t->∞
Note that e^t > t^2/2, and t/e^t > 0 for x > 0
0 < t/e^t < 2/t, lim 2/t = 0
t->∞
By The Squeeze Theorem
lim xlnx = 0
x->0+
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關於 e^t > t^2/2
令 f(x) = e^x - (x^2/2)
f'(x) = e^x - x > 0 for x > 0
故 f(x) 為一遞增函數, 且 f(0) = e^0 - 0 = 1 > 0
f(x) = e^x - (x^2/2) > 0 在 x > 0 恆成立
故知 e^x > x^2/2
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◆ From: 114.46.117.24
※ 編輯: BaBi 來自: 114.46.140.6 (11/20 11:42)