When are these two definitions equivalent? I can't remember how to show this but I feel like it must be true: if I have a continuous function $f$, then how do I show that 
$$\lim_{n\rightarrow \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}= c \quad\text{implies}\quad \lim_{\delta\rightarrow 0}\frac{f(x+\delta)-f(x)}{\delta}=c$$
EDIT: Sorry for the confusion, I did indeed mean $n\in \mathbb{N}$.
 A: This is not true. Consider 
$$f(x) = \begin{cases} x\cos\left(\frac {2\pi} x\right) &\text{if }x\neq 0,\\ 0 & \text{if }x=0.\end{cases}$$
Then $f$ is continuous on $\mathbb R$, 
$$\frac{f\left(\frac 1n\right) - f(0)}{\frac 1n} = 1$$
for all $n\in \mathbb N$, but $f$ is not differentiable at $0$. 
A: New Answer
I wrote this under the impression that $n$ and $\delta$ were both real, before the question was clarified. It is critical to the question that $n\in\mathbb{N}$, not $\mathbb{R}$, because then we can take advantage of periodic functions like sine or cosine to make the left-side limit exist. It is also critical that ${1\over\delta}\in\mathbb{R}$, not $\mathbb{N}$, because we can't take advantage of periodic functions unless they're discrete (which violates the "continuous function" requirement in the question).
Essentially, the left equation is no longer the definition of a derivative, though it would still serve just fine in most cases, because we're taking the limit discretely instead of continuously. But the right equation is the true derivative, with the limit taken continuously. So they aren't identical, and there are therefore ways to show implication is false, as shown by John Ma.
Old Answer
Basically, we can ignore most of the equation. The difference between the left side and the right side is that on the left we're taking the limit as ${1\over n}\to{1\over\infty}$ while on the right we're taking the limit as $\delta\to 0$.
So let's look at the limits directly.
$lim_{n\to\infty}{1\over n}$ can be solved by looking at a table or graph, and noting the value gets arbitrarily close to 0 as $n\to\infty$. There might be a more formal way to do it, but that's how we "proved" it in my calculus class.
$lim_{\delta\to 0}{\delta}$ can be solved by just plugging 0 into $\delta$ to get 0.
So we can let $\delta={1\over n}$ and note that as $n\to\infty$, ${1\over n}\to 0$ and $\delta\to 0$.
So once we've convinced ourselves that $\delta\to0$ is the same as ${1\over n}$, $n\to\infty$, the original implication makes sense. Start by noting that $n\to\infty\iff{1\over n}\to 0$:
$lim_{n\to\infty}{f(x+{1\over n})−f(x)\over{1\over n}}=c\iff$ $lim_{{1\over n}\to 0}{f(x+{1\over n})−f(x)\over{1\over n}}=c$
Now make the substitution, $\delta={1\over n}$:
$lim_{{1\over n}\to 0}{f(x+{1\over n})−f(x)\over{1\over n}}=c$ $\iff lim_{\delta\to 0}{f(x+\delta)−f(x)\over\delta}=c$
As John Ma's answer points out, it's possible to have a continuous function where the above limit doesn't exist for one or more $x$ values. It's actually possible to make a continuous function where the above limit exists nowhere. That's why we say a derivative only exists where the function is continuous and "smooth" or "well-behaved" near the point in question.
However, the implication doesn't say "the limit exists for all $x$". The implication says "if the limit of the left equation exists for a given $x$ and $f(x)$, then the limit of the right equation for the same $x$ and $f(x)$ not only exists, but has the same value, $c$".
We can go a step further by making it if and only if, as I've done above. If the limit exists for one side, it must exist for the other side. If the limit doesn't exist for one side, it can't exist for the other side. Because ultimately the two equations say the same thing.
