Re: Another question about derivatives

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

.

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