lim [(e^h-1)/h], h tends to 0
如果唔用differentiation (包括l'hospital, taylor...) 既話,
用first principle 要點做?
thx. |
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請問是1) y=[(e^h)-1]/h 還是2) y=[e^(h-1)]/h?
如果是1),
when h -> 0+, y -> +infinity
when h -> 0-, y -> -infinity
therfore the limit h --> 0 does not exist.
如果是2),
h --> 0, y=1
做法之一,plot graph!!!!! |
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[ref=329520]Kennethwu[/ref] 在 2005-8-6 02:10 PM 發表:
請問是1) y=[(e^h)-1]/h 還是2) y=[e^(h-1)]/h?
如果是1),
when h -> 0+, y -> +infinity
when h -> 0-, y -> -infinity
therfore the limit h --> 0 does not exist.
如果是2),
h - ...
係1, [(e^h)-1]/h
h -> 0+, y -> 1
h -> 0-, y -> 1
[ot 這是d/dx (e^x) 時出現的.
by def'n,
d/dx (e^x)
=lim [e^(x+h)-e^x]/h, h->0
=lim (e^x)(e^h-1)/h, h->0]
h=0, 分子=分母=0,
by l'hospital rule,
lim [(e^h)-1]/h, h->0
= lim e^h/1, h->0
=e^0
=1
但在未有differentiation 前,
人們用什麼方法証明y=1?
再次多謝. |
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