# Derivatives and Continuity

1. Jan 14, 2009

### JG89

1. The problem statement, all variables and given/known data

If the continuous function f(x) has a derivative f'(x) at each point x in the neighborhood of $$x=\xi$$, and if $$f'(x)$$ approaches a limit L as $$x \rightarrow \xi$$, then show $$f'(\xi)$$ exists and is equal to L.

2. Relevant equations

3. The attempt at a solution

Since the derivative exists at each point x in the neighborhood of $$x = \xi$$ and f'(x) tends to a limit L as $$x \rightarrow \xi$$, we have $$|f'(x) - L| < \epsilon$$ whenever $$|x-\xi| < \delta$$. Since we can take x arbitrarily close to $$\xi$$, we can write $$x = \xi + h$$ for any real h. In this case, we will take h to be positive. We then have $$|f'(\xi + h) - L| < \epsilon$$ whenever $$|h| < \delta$$.

Now, choose a positive quantity h* such that $$0 < h < h* < \delta$$. The following inequality is then true: $$\xi < \xi + h < \xi + h*$$. We have formed an open neighborhood about the point $$x = \xi + h$$. Then by the Mean Value Theorem, we have $$f'(\xi + h) = \frac{f(\xi + h) - f(\xi)}{h*}$$. So we then have $$|\frac{f(\xi + h) - f(\xi)}{h*} - L| < \epsilon$$ whenever $$|h*| < \delta$$. This inequality says that the limit of the quotient $$\frac{f(\xi + h) - f(\xi)}{h*}$$ as h* approaches 0 through positive values exists, and is equal to L. Using a similar argument it is easy to see that for h* tending to 0 through negative values, the limit is again L. Thus the derivative at $$f'(\xi)$$ exists and is equal to L.

Is this proof correct?

Last edited: Jan 14, 2009
2. Jan 15, 2009

### Dick

There's definitely some problems though I think you've got the right idea. The MVT doesn't say
$$f'(\xi + h) = \frac{f(\xi + h) - f(\xi)}{h*}$$
It says
$$f'(\xi + h) = \frac{f(\xi + h*) - f(\xi)}{h*}$$
And I think you should be able to write this much more simply, e.g. the limit h->0 of (f(xi+h)-f(xi))/h is f'(xi). The MVT says there is a point y_h in [xi,xi+h] such that f'(y_h) is the same as the difference quotient. As h->0, y_h->xi so f'(y_h)->L. And there's really no need to split it into sides, you don't have any one sided limits or derivatives.

3. Jan 15, 2009

### JG89

Oops, I mean't to write $$f(\xi + h) = \frac{f(\xi+h*) - f(\xi)}{h*}$$ for h* > h. I chose h* > h because if I chose h* < h, then I have xi < xi + h* < xi + h, and I cannot write out the derivative of xi + h as the difference quotient using xi and xi + h* (I believe the MVT says that xi + h must be contained in the open interval (xi, xi+h*).) But choosing h* so that h < h* < delta, h* goes to 0 and the quotient [f(xi + h) - f(xi)]/[h*] will go to L since the quotient - L is less than epsilon.

I see what you're saying though. I have made the problem more complicated than I had to. But with my correction above, is my proof correct?

Last edited: Jan 15, 2009
4. Jan 15, 2009

### Dick

I believe (f(x+h)-f(x))/h approaches f'(x). And I believe there is an 0<h*<h such that f'(x+h*)=(f(x+h)-f(x))/h. But I don't think (f(x+h)-f(x))/h* has to approach anything.

5. Jan 15, 2009

### JG89

But if you have 0 < h < h* then you have $$f'(\xi + h) = \frac{f(\xi + h*) - f(\xi)}{h*}$$ which approaches L as h* approaches 0. Since L is a definite value, then the limit exists. Isn't this the definition of a derivative or am I missing something here...

Last edited: Jan 15, 2009
6. Jan 15, 2009

### Dick

That's fine. As h*->0 the left side approaches L because the limit of the derivative exists and since h*>h>0 that means h->0 which means the right side approaches f'(xi) since f is differentiable at xi. That's the proof all right. I was just worried because you were writing expressions like (f(xi+h)-f(xi))/h*, mixing h and h* in the difference quotient.