Show that $\displaystyle f(x)$ is the derivative of$\displaystyle f(.)$ at$\displaystyle x$ if and only if $\displaystyle lim_{h \to 0} \sup_{|t|\leqslant h} \frac{|f(x+t)-f(x)-tf^{'} (x)|}{h} = 0 $

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- Oct 29th 2012, 11:50 AMujgilanihow to show that the following equality holds.
Show that $\displaystyle f(x)$ is the derivative of$\displaystyle f(.)$ at$\displaystyle x$ if and only if $\displaystyle lim_{h \to 0} \sup_{|t|\leqslant h} \frac{|f(x+t)-f(x)-tf^{'} (x)|}{h} = 0 $

- Oct 29th 2012, 10:53 PMchiroRe: how to show that the following equality holds.
Hey ujgilani.

Have you considered the triangle inequality? - Oct 29th 2012, 11:05 PMujgilaniRe: how to show that the following equality holds.
Hi, thanks for your reply

no I didn't try doing it with triangle inequality. could you please brief a little bit more. - Oct 29th 2012, 11:21 PMchiroRe: how to show that the following equality holds.
Basically the triangle inequality says |a+b| <= |a| + |b| where |.| is some norm.

- Oct 29th 2012, 11:24 PMujgilaniRe: how to show that the following equality holds.
yeah i know the "triangle inequality" concept but question is how we can apply it here

- Oct 29th 2012, 11:26 PMchiroRe: how to show that the following equality holds.
Show that |a| + |b| = 0.

- Oct 29th 2012, 11:33 PMujgilaniRe: how to show that the following equality holds.
sorry no idea how to use it in this setup

- Oct 29th 2012, 11:52 PMchiroRe: how to show that the following equality holds.
I think I may have given you the wrong advice: I'm thinking that since h > 0 (limit approached from the right) then |h| = h so you can put the h inside the absolute value sign. I'm sorry about the wrong advice.

Now one approach I am thinking of is to find a situation when you can interchange the limit signs within the absolute value from outside. I do know there are theorems that gaurantee when you can do this in some situations like this:

http://www.math.cuhk.edu.hk/course/1...onvergence.pdf