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I stumbled upon this Theorem here on pages 91 and 92 in the text. Check the complete Theorem there but it's basically this one. I am having trouble understanding two parts in the final step of the proof:

$(1)$ Now, as $x \rightarrow a$ every point in the interval $[a,x]$ gets arbitrarily close to $x$, so $$\lim_{x\to a}{c_x}=x$$ Wouldn't it be $\lim_{x\to a}{c_x}=a$ instead?

$(2)$ Thus, $$f'_R(a)=\lim_{x\to a^{+}}\frac{f(x)-f(a)}{x-a}=\bbox[yellow]{\lim_{x\to a^{+}}f'(c_x)=\lim_{x\to a^{+}}f'(x)}=L$$ I'm not quite sure I follow that highlighted part, can someone enlighten me?

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2 Answers2

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  1. You're absolutely right, it should be $\lim_{x \to a^+} c_x = a$.
  2. Fix $\epsilon > 0$. First, since $\lim_{x \to a^+} f^\prime(x) = L$, choose $\delta_0 > 0$, such that $\lvert f^\prime(x) - L \rvert < \epsilon$ whenever $0 < x-a < \delta_0$. Now, since $\lim_{x \to a^+} c_x = a$ (with $c_x > a$), choose $\delta > 0$, such that $0 < c_x - a < \delta_0$ whenever $0 < x - a < \delta$. What can you now conclude about $\lvert f^\prime(c_x) - L \rvert$ whenever $0 < x - a < \delta$?
Branimir Ćaćić
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Your proof and your notes are missing the relevance of this theorem, and I also think something has been lost in translation.

The implication is: Derivatives cannot have simple discontinuities. They can have discontinuities, just not simple discontinuities.

How does Spivak say it?

Suppose $f$ is continous at $a$ and $f'(x)$ exists at all points in an interval that includes $a$, except perhaps at $a,$ and $\lim_\limits{x\to a} f'(x)$ exists:

Then $f'(a) =\lim_\limits{x\to a} f'(x)$

$f'(a) \lim_\limits{h\to 0} \frac {f(a + h) - f(a)}{h}$

for sufficiently small $h>0, f$ is continuous in $[a,a+h]$ and differentiable in $(a,a+h)$ and the mean value theorem applies

There exists an $\alpha_h \in (a,a+h)$ such that $f'(\alpha_h) = \frac {f(a+h) - f(a)}{h}$

And $\alpha_h$ approaches $a$ as $h$ approaches $0.$

$\lim_\limits{h\to 0^+} \frac {f(a + h) - f(a)}{h} = \lim_\limits{\alpha_h \to a} f'(\alpha_h) = \lim_\limits{x \to a} f'(x)$

And a similar argument is made for the left hand limit.

It does not have the step $\lim_\limits{x\to a^+} f(c_x) = \lim_\limits{x\to a^+} f'(x)$ that was troubling you.

And now for something completely different. How does Rudin tackle the problem? (Actually after including this I see that it is already in your book)

Rudin proves something that I think is a little bit more general.

If the derivative is defined everywhere on an interval, the intermediate value property applies.

Suppose $f$ is a differentiable on $[a,b]$ and suppose $f'(a) < \lambda < f'(b)$ then there exists an $x \in (a,b)$ such that $f'(x) = \lambda$

Consider $g(t) = f(t) - \lambda t$

As $g'(a) < 0$ then there is a $t_1> a$ such that $g(t_1) < g(a)$ and $g'(b) > 0$ implies there exists a $t_2 < b$ such that $g(t_2) < g(b)$

$g(t)$ attains a minimum at some point $x\in (a,b)$

$g'(x) = 0\\ f(x) = \lambda$

If $f'(a) > f'(b)$ a similar result holds.

Doug M
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