Re: limit of this function?

From: Thomas Mautsch (mautsch_at_math.ethz.ch)
Date: 07/16/04 

Date: 16 Jul 2004 18:35:35 +0100

In news:<20040716.061427@whim.org> schrieb Rob Johnson <rob@trash.whim.org>:
> In article <baa549eb.0407152253.6330acb@posting.google.com>,
> max1974isome@hotmail.com (Max) wrote:
>>Brute force calculating gives the limit of
>>
>> h
>>3 - 1
>>-------
>> h
>>
>>as approximately 1.09 when h approaches 0.
>>
>>How can I find that algebraically without
>>calculus?
>>
>>Using the squeezing theorem I was able to
>>confirm its range between 1 and 1.2817
>>(= 4-e). But that's as close as I can get.
>>
>>any ideas for an exact answer?
>
> Let us start with the original definition for e:
>
> n
> e = lim (1 + 1/n)
> n->oo
>
> Setting x = 1/n, we have
>
> 1/x
> e = lim (1 + x)
> x->0+
>
> Since 3 = e^ln(3), we get
>
> ln(3)/x
> 3 = lim (1 + x)
> x->0+
>
> Setting y = x/ln(3), we have
>
> 1/y
> 3 = lim (1 + y ln(3))
> y->0+
>
> That means that as y -> 0,
>
> y 1/y 1/y
> lim (3 ) - (1 + y ln(3)) = 0
> y->0+
>
> If we now look at y = 1/m, we have
>
> y m m
> 0 = lim (3 ) - (1 + y ln(3))
> y->0+
>
> y y(m-1) m-1
> = lim [3 - (1 + y ln(3))] [3 + ... + (1 + y ln(3)) ]
> y->0+ \________________________________/
> m terms, all -> 3
>
> y 3
> = lim [3 - (1 + y ln(3))] -
> y->0+ y
>
I don't want to say that this part is invalid,
but I just don't like it, because it gives the impression
that m is independent of the variable y, which at the same time tends to 0+.

> Therefore, moving the pieces of the preceding equation around, we get
>
> y
> lim (3 - 1)/y = ln(3)
> y->0+
>

Wouldn't it have been much shorter
to derive

        e^x = 1 + x/1! + x^2/2! + ...

(or even simpler

        e^x = 1 + x/1! + O(x^2) )

from your definition of e from above? -- When you think of it,
all you need to do then
is to set x = h*ln(3) in this formula, and you're done!

(This seems to be what you are doing,
 but you are introducing in the ln(3)-factor at a conceptionally
 too early point, I think.)

> To handle the other direction, let z = -y, then we can conclude
>
> y
> lim (3 - 1)/y
> y->0-
>
> -z
> = lim (1 - 3 )/z
> z->0+
> z -z
> = lim (3 - 1)/z 3
> z->0+
>
> = ln(3)
>
> since 3^{-z} -> 1.

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