What's the difference between the two？

Question

Question

Which of the following two can define the derivative $f'(a)$:

1)$$\lim_{n \to \infty}n \left[f\left(a+\frac{1}{n}\right)-f(a)\right],n \in \mathbb{Z}.$$

2)$$\lim_{x \to \infty}x\left[f\left(a+\frac{1}{x}\right)-f(a)\right],x \in \mathbb{R}.$$

Here, the fact $\lim\limits_{x \to \infty}$ exists implies that $\lim\limits_{x \to +\infty}$ and $\lim\limits_{x \to -\infty}$ both exist and equal. — mengdie1982, Nov 15 '18 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative. — Paramanand Singh, Nov 15 '18 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh — mengdie1982, Nov 15 '18 at 14:36
Refer to this answer for substitution rule of limits: https://math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity. — Paramanand Singh, Nov 15 '18 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values. — Paramanand Singh, Nov 15 '18 at 14:41
According to 2) the absolute value function would be differentiable at $0$. — Christian Blatter, Nov 15 '18 at 15:40

score 2 · Answer 1 · edited Nov 15 '18 at 14:54

Let $f(x)= \begin{cases} 0 & \text{If $x$ is rational.} \\ 1 & \text{If $x$ is irrational} \end{cases}$

Then $\displaystyle \lim_{\substack{n \to \infty} \\ n \in \mathbb{Z}}n \left[f\left(0+\frac{1}{n}\right)-f(0)\right] = 0$

and

$\displaystyle \lim_{\substack{n \to \infty} \\ n \in \mathbb R \setminus \mathbb{Q}}n \left[f\left(0+\frac{1}{n}\right)-f(0)\right] = \infty$.

Hence

$\displaystyle \lim_{\substack{n \to \infty} \\ n \in \mathbb R}n \left[f\left(0+\frac{1}{n}\right)-f(0)\right]$ does not exist.