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标题: limit [打印本页]

作者: yma16 时间: 2007-4-27 06:52
标题: limit

As I remember, limit of x^x when x goes to 0+ is 1. What is the limit of

when x goes to 0+

作者: yinyin 时间: 2007-4-27 07:58
标题: 回复：limit

Positive infinite.

作者: yma16 时间: 2007-4-27 08:00
标题: 回复：回复：limit

Maybe not.

作者: yinyin 时间: 2007-4-27 08:03
标题: 回复：回复：回复：limit

For each x>0, it is positive infinite. So, when x goes to 0+, its limit is positive infinite too.

作者: yma16 时间: 2007-4-27 19:12
标题: 回复：回复：回复：回复：limit

When x=1, it is 1. If you search my old post, you can see someone found the max point which is x=e^(1/e).

作者: yinyin 时间: 2007-4-27 23:57
标题: Sorry[:>]

Sorry, I thought that x goes to +1 but wrote+0. I am so stupid!

Let me think it again.

作者: yinyin 时间: 2007-4-28 01:08
标题: 回复：回复：回复：回复：回复：limit

The limit should be 1. For any given x>0 (but <1), the value of your expression is 1. So, when x goes to 0+, its limit is 1 too.

Please note that, there are two limits in such kind of problems: one is "the number, n, of x's in the expression goes to infinite", another is "x goes to 0+". Now, in the given problem, the former is taken first, then the latter is taken. I guess that value e^(1/e) may be reached when the above-mentioned two limits are taken at the same time according to a certain relation between n and x.

作者: calala 时间: 2007-4-28 03:51
标题: 回复：limit

Let y = lim_x->0+ x^x^x^...

y^x = y

y^(1-x)=1

lim_x->0+ y^(1-x) = lim_x->0+ y^1 = 1

==> y = 1

本贴由[calala]最后编辑于：2007-4-27 20:50:49

作者: yinyin 时间: 2007-4-28 06:47
标题: 回复：回复：limit

There is a problem in your proof. How can you obtain the second equality from the first one?

In fact, from the second equality, we may obtain y=1 immediately.

作者: yma16 时间: 2007-4-28 18:17
标题: 回复：回复：limit

It seems you can use this way to show any limit to be 1. For example, when x goes to 2, y is 1. But you know when x=2, y is infinity.

作者: yma16 时间: 2007-4-28 18:27
标题: 回复：回复：回复：回复：回复：回复：limit

You said "For any given x>0 (but <1), the value of your expression is 1." This is not true.

Let y be the expression without the limit (i.e. y=x^x^x...). Then y=x^(x^x...)=x^y. According to you, when x=.5, y=1. but 1<>.5^1.

作者: yinyin 时间: 2007-4-28 22:46
标题: [:-M]

看了yma兄的新帖，才突然明白问题的所在：所给表达式是有歧意的。你我对表达式的含义理解不一，当然就说不到一块去了。清查看yinyin对yma兄新帖的跟帖，其中指出了为什么yma兄的数学表达式是不严谨的。

yinyin对yma兄的数学表达式的理解是：当x的个数趋于无穷时(...((x^x)^x)...)^x的极限。请yma兄解释一下你对该表达式的定义。www.ddhw.com

作者: yma16 时间: 2007-4-29 02:17
标题: 回复：[:-M]

The lean tower exponent is a standard math expression. x^x^x... is confusing but it is easier to write with PC. As I posted on the other limit question, you compute it from the top to the bottom.

作者: yinyin 时间: 2007-4-29 05:07
标题: 回复：回复：[:-M]

yinyin还是没有看出yma兄是怎样定义该表达式（有无穷多个x）的。请明示。

作者: yma16 时间: 2007-4-29 20:31
标题: 回复：回复：回复：[:-M]

Let a[0]=x, ..., a[n+1]=x^a[n]...

The limit is the 二重極限

limit a[n] when x goes to 0+ and n goes to infinity.

作者: yinyin 时间: 2007-4-29 23:31
标题: 回复：回复：回复：回复：[:-M]

如果就把此极限作为那表达式的定义，其极限值记为y(依赖于x)，那还存在两个问题：

１）如何证明y=x^y?;

２）按原问题，应是累次极限，即先让n趋于无穷，然后让x趋于0+。若考虑成二重极限，那么极限值往往会跟取极限的方向（在那两个变量构成的二维空间中）有关。

作者: yma16 时间: 2007-4-30 02:35
标题: 回复：回复：回复：回复：回复：[:-M]

1. Assume you have an expression y=xxx...,

(a) I think you can write y=x(xxx...) =>y=xy.

(b) If you have another expression z=0.1xxx... you can write z=.1y.

If you agree with either (a) or (b), you should agree with y=xy when y=x^x^x...

2 累次极限 is OK. I think they should be the same.

作者: yinyin 时间: 2007-4-30 04:09
标题: 回复：回复：回复：回复：[:-M]

作为这样一个无穷多阶次指数函数的表达式，其定义要有一般性，即如果表达式中这可列多个x从下到上分别被非负数列 a₁,a₂,a₃,......（或其他具有某种一般性的非恒等数列）代替，这表达式仍然有确定意义。请考虑对上述非负数列，a₁^a₂^a₃^......是什么意思。如果该表达式的定义仅对恒等数列有效，那似乎就应考虑有没有必要引进这么一个表达式。直接写成f(x)，并定义它为方程f(x)=x^f(x)的解就行了。

yinyin先前给出的此表达式的另一种定义，就具有一般性。

作者: yma16 时间: 2007-4-30 21:06
标题: 回复：回复：回复：回复：回复：[:-M]

方程f(x)=x^f(x) is fine. But it will not help us to find the limit.

作者: yinyin 时间: 2007-5-1 03:54
标题: 回复：回复：回复：回复：回复：回复：[:-M]

Whose "limit"? How will you find the "limit"?

First, we should have a proper definition for the relevant expression. Then, we may consider the limit.

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