Define the operator $\Delta_n$ according to the equation
$$\Delta_nf(x)=f\left(x+\frac1n\right)-f(x)$$
Observe that for differentiable $f:\Bbb{R}\to\Bbb{R}$
$$\frac{df}{dx}=\lim_{n\to\infty}n\Delta_nf$$
(Note: The limit can be evaluate from either side by changing the sign of $n$)
This matters solely because it is easier to prove that the sequence $(n\Delta_nf)_{n\in\Bbb{N}}$ converges to some limit $L$ than it is to prove that $\lim_{h\to0}(f(x+h)-f(x))/h=L$ over the reals – so much so, that it's tempting to use this as the definition of the derivative.
So why isn't this the definition of the derivative?
The most significant reason that I can think of is that while the existence of the derivative implies the above equation the converse does not hold. It is possible to have a function such that the above sequence converges when the derivative does not exist. For example, take:
$$g(x)=\begin{cases}e^x & x\in\Bbb{Q}\\0 & \text{otherwise}\end{cases}$$
The sequence $n\Delta_ng(x)$ converges to $g(x)$ for all $x$, but $g$ is not continuous – hence, not differentiable – at any point of its domain.
This problem is easily resolved by adding the qualification "if $f$ is continuous at $x$," since this is a relatively simple condition to check in many cases. So the new definition of the derivative is as follows:
For a function $f:E\subseteq\Bbb{R}\to\Bbb{R}$, continuous at a point $x\in E$, the derivative of $f$ at $x$ exists and is equal to $\lim_{n\to\infty} n\Delta_nf(x)$ iff the sequence $(n\Delta_nf(x))_{n\in\Bbb{N}}$ is convergent.
This sounds correct, but it still leaves the possibility of pathological counterexamples. Continuous nowhere-differentiable functions come to mind, but for every example I can think of, the above sequence does not converge.
Are there any examples of a continuous, non-differentiable function s.t. $\lim_{n\to\infty} n\Delta_nf$ still converges?
Best Answer
Let $f(x) = x\sin(\pi/x)$ if $x \neq 0$ and set $f(0) = 0$.
Then $f$ is continuous but not differentiable at the origin. But
$$ \Delta_n f(0) = \frac{\sin(\pi n)}{n} = 0, $$
so the rational derivative exists and is zero.