Can we use this as definition of the derivative? Let $D$ be a subset of $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $f:D\to \mathbb{K}^N$ be a function. Then $f$ is differentiable at $x$ if $$\lim_{n\to \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}$$ exists.
 A: No, you cannot. At least this definition is not equivalent to the usual definition of the derivative. Let's take an example - in your setting we have $D=\mathbb K=\mathbb R$ and $N=1$:
$$
f(x):=\begin{cases} \sin(\frac \pi x), x\neq0\\ 0,x=0 \end{cases}
$$
then we have in $x=0$
$$
\lim_{n\to\infty}\frac{f(0+\frac{1}{n})-f(0)}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\sin(n\pi)-0}{\frac 1 n}=0
$$
so the derivative would be $0$ at $x=0$. But in fact $f(x)$ is not even continuous at $x=0$ and therefore for sure not differentiable.
Maybe an even more apparent example would be the modulus function
$$
g(x)={\displaystyle |x|={\begin{cases}\ \;\,\ \ x&{\text{für }}x\geq 0\\\ \;\,-x&{\text{für }}x<0\end{cases}}}
$$
here you see right away, that your definition would give us the derivative of $g(x)$ in $x=0$ to be $1$, since you'r oppressing the limiting sequence coming from the negative side. This clearly conflicts with the usual definition.
As @GEdgar already mentioned in one comment, we have nevertheless the implication, that if any function is differentiable, then it will also satisfy your limit process (with the same limit) (and if its not satisfied, then the function is not differentiable in the usual sense). 
So your definition is a special case of the usual definition.
