Does the Limit of f'(x) as x Approaches xi Guarantee f'(xi) Equals L?

[JG89]

Homework Statement

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.

Homework Equations

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?

 

[Dick]

There's definitely some problems though I think you've got the right idea. The MVT doesn't say
<br /> f&#039;(\xi + h) = \frac{f(\xi + h) - f(\xi)}{h*}<br />
It says
<br /> f&#039;(\xi + h) = \frac{f(\xi + h*) - f(\xi)}{h*}<br />
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 anyone sided limits or derivatives.

 

[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?

 

[Dick]

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?

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.

 

[JG89]

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.

But if you have 0 < h < h* then you have f&#039;(\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...

 

[Dick]

But if you have 0 < h < h* then you have f&#039;(\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...

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.