Cont. & almost everywhere diff => derivative exists?

[aswilder]

I am doing some self-study of basic real analysis. While for the most part the exercises in the text [A First Course in Real Analysis, by M.H. Protter, 2nd edition] to me are fairly straight-forward (in days gone by I was a math major), I find the following exercise in the introductory chapter on derivatives in R[sup:3odcqg48]1[/sup:3odcqg48] has me positively stumped. I seem to keep circling back to the same issue in trying to work out a proof.

Here is the exact text of the exercise (section 4.1, item 12.):

Suppose that f is continuous on an open interval I containing x[sub:3odcqg48]0[/sub:3odcqg48], f' is defined on I except possibly at x[sub:3odcqg48]0[/sub:3odcqg48], and f'(x) approaches L as x approaches x[sub:3odcqg48]0[/sub:3odcqg48]. Prove that f'(x[sub:3odcqg48]0[/sub:3odcqg48]) = L.

When I draw a picture, I agree with the proposition. Namely since f is continuous it has no gaps at x[sub:3odcqg48]0[/sub:3odcqg48]. The fact that f'(x) approaches L on either side seems to suggest no kinks are present. But every inequality I write seems to pertain to the situation just to the left or right of x[sub:3odcqg48]0[/sub:3odcqg48]. What I need is to do is to zoom in on the value {f(x[sub:3odcqg48]0[/sub:3odcqg48]+h)-f(x[sub:3odcqg48]0[/sub:3odcqg48])}/h for small h. Any help or hints would be fantastic. I have been struggling with this problem off and on for many hours. Thanks.

 

[daon]

If you can use L'Hopital's Rule, then

\(\displaystyle \lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} = \lim_{h \to 0}\frac{f'(x_0+h)}{1}\)

If we let \(\displaystyle x = x_0+h\) then

\(\displaystyle \lim_{h \to 0}\frac{f'(x_0+h)}{1} = \lim_{x \to x_0}f'(x) = L\)


If not, the proof will likely be very tedious. You'll need to show:

\(\displaystyle (\forall \epsilon > 0)(\exists \delta > 0), \,\, x \in (x_0-\delta, x_0) \cup (x_0, x_0+\delta) \implies L - \epsilon < \lim_{h \to 0} \frac{f(x_0+h) - f(x)}{h} < L + \epsilon\)

If you must go this route I would assume \(\displaystyle f'(x_0) \neq L\) to try to contradict that \(\displaystyle \lim_{x \to x_0}f'(x) = L\).

 

[aswilder]

Yes, that's it. L'Hopital's rule is the trick. In fact that is covered in the same section of this textbook. The key is that this rule does not require that when you differentiate the numerator and denominator, there is no requirement that individually the derivatives exist at the limit a at which you are tyring to evaluate. Just that the ratio exist in the limit.

Piece of cake once you see the trick and think of the issue in this way. Thanks ever so much.