Re: need with my sons homework please

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 09/28/04 

Date: Tue, 28 Sep 2004 16:08:35 GMT

Geert van der Wulp wrote:
> On 28 Sep 2004, Geert van der Wulp wrote:
>
>>On 28 Sep 2004, Andreas wrote:
>>
>>>Hello everyone and greetings form cold sweden
>>>
>>>I have a terrible problem with my sons homework and since I do really
>>>wanna help him im turning to you guy for help.
>>>
>>>The problem is that hes got to calculate (Im from sweden so hope my
>>>english disturbe you to much :))the following:
>>>"calculate the function e^h-1/h when h moves towards 0" and the other
>>>one:
>>>"calculate (1-x^2)/(1-x) when x moves towards 1"
>>>
>>>Now what in gods name is all this? I know ordinary algebra but this?
>>>Ive searched the net and found that the function isnt defined for the
>>>number 0 and 1 and some examples but still cant get it right! any
>>>suggestions or solutions would be greatly appriciated.
>>>
>>>Thanks
>>>Andreas R
>>>
>>>Ive also posted this message to you guys since the swedish groups are
>>>reaaaally slow to answer.
>>
>>Hi Andreas,
>>
>>Let us start with the second question. 1-x^2 = (1-x)(1+x)
>>So lim x-->1 (1-x^2)/(1-x) = lim x-->1 (1-x)(1+x)/(1+x) = 0
>>
>>For the second question it is not clear to me if you needed to put brackets or not. So let me split this in two cases:
>>a) lim h-->0 e^h - 1/h = e^0 - lim h-->0 1/h = minus infinity
>>b) lim h-->0 e^(h-1/h) = lim h-->0 (e^h * e^(-1/h) =
>>lim h-->0 e^(-1/h) = 0. (Because e^-infinity := 0, please this is no mathematically accepted notation, only to understand)
>>c) lim h-->0 (e^(h-1))/h = lim h-->0 e^(-1)/h = infinity
>>
>>I hope this solves your questions?
>>
>>With kind regards,
>>
>>Geert
>
>
> Hmmm, way too quick of course.
>
>
>>a) lim h-->0 e^h - 1/h = e^0 - lim h-->0 1/h = minus infinity
>
>
> or plus infinity, depending on the sign of h
>
>
>>b) lim h-->0 e^(h-1/h) = lim h-->0 (e^h * e^(-1/h) =
>>lim h-->0 e^(-1/h) = 0
>
>
> Or infinity, depending on the sign of h
>
>
>>c) lim h-->0 (e^(h-1))/h = lim h-->0 e^(-1)/h = infinity
>
>
> Or minus infinity, depending on the sign of h,
>
> I am beginning to wonder if I did understand your first function correctly.
>
> Geert
>

Everyone seems to realize that the function in
the first problem is rendered ambiguously, and
several people have offered their renditions of
what that function ought to have been.

The correct parsing of this

        e^h-1/h

should be as follows

        (e^h - 1)/h

I'm astonished that no one seems to have seen it,
especially considering that many folks recognized
the source as an introductory calculus class.

Dale.

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