Is function of convergent sequence rate of change equal to derivative?

[brian44]

Given a convergent sequence $$x_n \rightarrow x$$ and a function f, is
$$lim_{n \rightarrow \infty} \frac{f(x_n) - f(x_{n-1})}{x_n - x_{n-1}} = f'(x)$$ ?

I believe it it is, but I haven't been able to figure out how to prove it. Does anyone know of a proof or counter-example?

And probably should add $$x_n \not = x_{n-1} \forall n$$

 

[HallsofIvy]

Saying that
$$\lim_{x\to a} f(x)= F$$
is exactly the same as saying
$$\lim_{n\to\infty} f(x_n)= F$$
where
$$\lim_{n\to\infty} x_n= a$$

so, yes, your equation is correct.

 

[Office_Shredder]

Certainly if the derivative is not continuous this isn't going to work. For the function f(x)=x²sin(1/x) when x is not 0, f(0)=0, this is differentiable with a derivative of 0 at 0. But its derivative is 2xsin(1/x)-cos(1/x) is not continuous (see wolfram alpha for a graph) and you can find a sequence of points converging to zero such that the difference quotient evaluates to whatever you want it to be really

 

[brian44]

We may assume f and f' are continuous, since that is not the main question I am concerned with.

Saying that
$$\lim_{x\to a} f(x)= F$$
is exactly the same as saying
$$\lim_{n\to\infty} f(x_n)= F$$
where
$$\lim_{n\to\infty} x_n= a$$

so, yes, your equation is correct.

It is not quite the same thing, since that is not how the derivative is defined - all definitions I've seen have one value fixed. I.e. just plugging in $$x_n$$ to $$f'(x)$$ does not give the same function as I described, so sequential characterization doesn't work directly:

$$f'(x_n) = \lim_{x \rightarrow x_n} \frac{f(x) - f(x_n)}{x-x_n}$$

or we could define it as

$$f'(x) = \lim_{n \rightarrow \infty} \frac{f(x) - f(x_n)}{x-x_n}$$

However neither of these are the same as the limit I gave, in which both $$x_n$$ and $$x_{n-1}$$ are changing sequences, and aren't fixed. Even if accepted as another definition, then can you offer a proof the definitions are equivalent?The problem I was stuck with when I tried to use something like
$$f'(x_n) = \lim_{x \rightarrow x_n} \frac{f(x) - f(x_n)}{x-x_n}$$
to get a bound, is that then any delta requirement depends on the specific value of n, but I may need to choose n large enough so that $$x_n$$ and $$x_{n-1}$$ meet the delta requirement, but that ends up potentially changing $$x_n$$ and the delta needed again since it depends on the specific x_n, so it's sort of like a race condition.

 

[snipez90]

Well if f' is continuous at x and f' is defined in a neighborhood of x, you should be able to apply the mean value theorem to the interval consisting of the endpoints x_(n-1) and x_n (taking n to be large enough to begin with). Then letting n approach infinity and using the continuity of f' at x seems to establish the desired equality.

Note that I did in fact consider Office_Shredder's function (since it is more or less the canonical example of a function whose derivative exists everywhere but does not have a continuous derivative), and for obvious sequences tending to x = 0 I think the equality still holds.

 

[brian44]

Well if f' is continuous at x and f' is defined in a neighborhood of x, you should be able to apply the mean value theorem to the interval consisting of the endpoints x_(n-1) and x_n (taking n to be large enough to begin with). Then letting n approach infinity and using the continuity of f' at x seems to establish the desired equality.

Note that I did in fact consider Office_Shredder's function (since it is more or less the canonical example of a function whose derivative exists everywhere but does not have a continuous derivative), and for obvious sequences tending to x = 0 I think the equality still holds.

Ah, thanks! I didn't even think of using the mean-value theorem...