Re: Another question about derivatives
- From: quasi <quasi@xxxxxxxx>
- Date: Mon, 26 Dec 2005 21:25:36 -0500
On 26 Dec 2005 18:05:14 -0800, "Dave L. Renfro" <renfr1dl@xxxxxxxxx>
wrote:
>The World Wide Wade wrote:
>
>>> I see we're having a little too much eggnog
>>> during the holidays. [f(x)-f(0)]/(x-0) = f'(c_x),
>>> right? So if f'(c_x) -> 0, then ...
>
>quasi wrote:
>
>> then lim f'(x) --> 0 as x --> 0.
>>
>> What am I missing?
>
>Well, actually you don't know that f'(x) --> 0 as x --> 0
>simply from f'(c_x) --> 0 (the stronger limit, where x --> 0,
>might not exist, while the weaker limit, where x approaches
>zero in such a way that x is always equal to one of the c_x's,
>might exist), although of course we do know that f'(x) --> 0
>because this was in the original poster's hypothesis. [I'm not
>suggesting that you didn't recognize this as in the original
>hypothesis, by the way. Indeed, I assume this was part of
>your confusion, namely that the argument seems to only prove
>something we already knew. My point is that the argument
>doesn't even do this as far as I can tell.]
>
>I believe what's going on is that in the equation
>
> [f(x)-f(0)] / (x-0) = f'(c_x),
>
>if we take the limit as x --> 0 of both sides, then the
>left side becomes the definition of f'(0) and the right
>side becomes 0.
>
>Dave L. Renfro
Thanks, Dave.
I did finally realize what I was missing.
quasi
.
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