Re: derivatives as limit of a sequence of continuous functions

Artur wrote:

I'm trying to prove the following fact: if f:I --> R is differentiable
on an open interval I, then f' is the limit of a sequence of continuous
functions defined on I.

First, I considered an interval like (a, oo) with a real or -oo. Let
t_n be a sequence of positive numbers that converges to 0 and, for
every positive integer n, define g_n(x) = (f(x + t_n) - f(x))/(t_n).
Then, it's easy to see that each g_n is continuous and g_n -> f'. For
intervals like (-oo, a) the proof is similar.

A trickier case happens if we have intervals like (a,b), with a and b
real. Admitting f has a limit L at b, we can suppose t_n is in (0, b-a)
for every n and define g_n(x) = (f(x + t_n) - f(x))/(t_n) if  a < x < b
- t_n  and g_n(x) = (L - f(b -t_n))/(t_n) if b-t_n <= x < b. Then each
g_n is continuous on (a, b) and, for sufficient large n, every x in n
(a,b) will satisfy x < b - t_n, so that g_n => f'.

But I couldn't handle the case (a,b), and b real, if f doesn't have a
limit at a nor at b. In this case, the previous reasonings donn't work
and I'd like a suggestion. 

I suggest that your t_n's become functions, instead of being constant.

Best regards,

Jose Carlos Santos
.

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