A Different Definition of Derivative

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Homework Statement



part 1. )

If f:(a,b)--> R is differentiable at p in (a,b), prove that

f ' (p) = lim (n -> oo) n [ f(p + 1/n) - f(p) ].

part 2. )

Show by example that the existence of the limit of the sequence,

{n [ f(p + 1/n) - f(p) ] } does not imply the existence of f ' (p).

Homework Equations





The Attempt at a Solution



One failed attempt involved comparing this limit to the definition below,

\lim_{n \rightarrow \infty} \frac{f(p_n)-f(p)}{p_n-p}.

I attempted to break the new definition (up top) into the sum of the limits and proceed down that path but it did not lead to anything.

I have also tried this new definition with a functions and a point and compared its result to f'(p) to gain some understanding. I still cannot seem to get anywhere though.

Any help would be appreciated. Thank you.
 
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That sequence looks EXACTLY like your definition where p_n=p+1/n. What's the problem?
 
I see how this works for the sequence p_n= p + 1/n. Do we not have to show that the two are equivalent for all sequences that converge to p, and/or is that what we are doing?
 
If the function is differentiable at p then all such sequences converge to the derivative. For the second part they want to find a function that is NOT differentiable at p, but where the limit of that particular sequence does exist.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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