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.
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For $N = 1$, sure. For $N \ge 2$, you'll need to be more careful. – Alexis Olson Sep 06 '16 at 11:13
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2Suppose $f(0)=f(1/n)=17$ for all integer $n$, $f(x)=42$ for all other $x$. Then your limit exists at $x=0$, but your function's not even continuous there. – Gerry Myerson Sep 06 '16 at 11:20
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1Also, over $\mathbb{C}$ you are only taking the limit along the reals, which also causes problems. – user347489 Sep 06 '16 at 11:29
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3What you can say, however: if the derivative exists, then this limit computes it. – GEdgar Sep 06 '16 at 11:34
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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.
user190080
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Are you sure about your last limit ? $lim_{n \rightarrow \infty} sin(n \pi)/(1/n) = undef$ ? – Zubzub Sep 06 '16 at 11:24
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1@Zubzub actually...I am sure: $\forall n:\sin(n\pi)=0$ and therefore $\displaystyle \frac{\sin(n\pi)}{\frac 1 n}=n\sin(n\pi)=0 \implies \lim_{n\to\infty}n\sin(n\pi)=0$ – user190080 Sep 06 '16 at 11:27
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