作者LuisSantos ( )
看板trans_math
标题Re: [考古] 清大
时间Thu Jun 5 11:57:59 2008
※ 引述《le5868ov (我一定要上榜！~!~)》之铭言：
: Find
: 2 2
: -x x t
: lim x e ∫ e dt = ?
: x→∞ 0
2 2
-x x t
lim (x)(e )(∫ e dt)
x→∞ 0
2
x t
(x)(∫ e dt)
0
= lim --------------------
x→∞ e^(x^2)
x
∫ e^(t^2) dt + (x)(e^(x^2))
0
= lim --------------------------------
x→∞ (2x)(e^(x^2))
e^(x^2) + e^(x^2) + (2x)(e^(x^2))
= lim -------------------------------------
x→∞ (2)(e^(x^2)) + (2x)(2x)(e^(x^2))
(2)(e^(x^2)) + (2x)(e^(x^2))
= lim ----------------------------------
x→∞ (2)(e^(x^2)) + (4)(x^2)(e^(x^2))
(2)(e^(x^2))(1 + x)
= lim ----------------------------
x→∞ (2)(e^(x^2))(1 + (2)(x^2))
1 + x
= lim --------------
x→∞ 1 + (2)(x^2)
(1/(x^2)) + (1/x) 0 + 0 0
= lim ------------------- = ------- = --- = 0
x→∞ (1/(x^2)) + 2 0 + 2 2
: 1/x
: lim (1 + sin2x) = ?
: x→0
: 请多指教><""
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