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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 $n\in \mathbb{N}$.

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  • $\begingroup$ Good question! Never really understood this myself. However, you'll want not only "implies" but also "is implied by." $\endgroup$ – goblin Nov 2 '15 at 0:29
  • $\begingroup$ Sure, but that direction seems a bit more straightforward to me. $\endgroup$ – user12344321 Nov 2 '15 at 0:30
  • $\begingroup$ If f is continuous, the for all delta there is 1/(n+1)<= d <= 1/n and for all n there are delta_1 < 1/n < delta_2. So if one is true the other is. $\endgroup$ – fleablood Nov 2 '15 at 0:33
  • $\begingroup$ I expect that is the case, but how to do you actually use continuity to control what is happening? $\endgroup$ – user12344321 Nov 2 '15 at 0:36
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    $\begingroup$ If $n$ should be restricted to integers, it's worth clarifying that explicitly. $\endgroup$ – Hurkyl Nov 2 '15 at 9:36
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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 $0$.

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    $\begingroup$ Ah wonderful, thank you. $\endgroup$ – user12344321 Nov 2 '15 at 0:40
  • $\begingroup$ However, this would not work if we had initially assumed that for every sequence going to zero, the limit existed, correct? $\endgroup$ – user12344321 Nov 2 '15 at 0:42
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    $\begingroup$ @user12344321 : If you assume that limit exists for all sequence going to zero, then the limit has to be the same for all sequence, and in that case the limit actually exists, and your claim will be true in that case. $\endgroup$ – user99914 Nov 2 '15 at 0:44
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    $\begingroup$ From the comment above by the OP, I would assume that the OP is really talking about sequence and $n\in \mathbb N$. But, again, it's up to the OP to clarify. @PyRulez $\endgroup$ – user99914 Nov 2 '15 at 1:45
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    $\begingroup$ JohnMa is correct. $\endgroup$ – user12344321 Nov 2 '15 at 3:56
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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.

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