Question about defn. of derivative

[AxiomOfChoice]

Let $\{h_n\}$ be ANY sequence of real numbers such that [itex]h_n \neq 0[/tex] and $h_n \to 0$. If $f'(x)$ exists, do we have<br /> <br /> $$f'(x) = \lim_{n\to \infty} f_n(x),$$<br /> <br /> where <br /> $$f_n(x) = \frac{1}{h_n} (f(x+h_n) - f(x))$$<br /> <br /> ?<br /> <br /> This seems to express the derivative as the pointwise limit of a sequence of functions...right? Do we know, in addition, that the convergence is uniform?[/itex]

 

[mathman]

Uniform in what sense? The definition does not require uniformity in x.

 

[AxiomOfChoice]

Uniform in what sense? The definition does not require uniformity in x.

Ok, I'm not sure :) I didn't really think before writing out that question. But am I right on all other counts?

 

[mathman]

You are right, but it is more cumbersome than the usual approach where:

f'(x)=lim(h->0) (f(x+h) - f(x))/h

 

[HallsofIvy]

$lim_{x\to a} f(x)= L$ if and only if $\lim_{n\to \infty} f(a_n)= L$ for every sequence $\{a_n\}$ that converges to a. The two formulations are equivalent.