# Another corollary of the Lagrange's Mean Value Theorem

I'm having trouble at proving the following theorem, stated as a corollary of the Lagrange's Mean Value Theorem in the book I'm reading ("Curso de Análise", Elon Lages Lima):

Let $f: \left(a, b\right) \rightarrow \mathbb{R}$ be a differentiable function except, possibly, at $c \in \left(a, b\right)$, with $f$ continuous at $c$. If the limit:

$$\lim_{x \rightarrow c} f'(x)=L$$

exists, then $f'(c)$ exists and its value is $f'(c)=L$.

My attempt at proving this was:

"As long as $c \neq a$ and $c\neq b$, then there exists $A \in \left(a, c\right), B \in \left(c, b\right)$ such that $f$ is continuous at $(A, c)\cup(c,B)$ (because $f$ is differentiable in these open intervals), $f$ is continuous at $A$, $B$ and $c$ (because $f$ is differentiable at $A$ and $B$, when the intervals under consideration are $(a,c)$ and $(c,b)$ and $f$ is supposedly continuous at $c$). Then $f$ is (uniformly) continuous in $\left[A,B\right]$. In particular, $f$ is uniformly continuous in every closed interval contained in $\left[A,B\right]$.

Let $\epsilon > 0$ be given, and let's keep fixed a corresponding $\delta > 0$ which satisfies the uniform continuity. Then $f$ is (uniformly) continuous at $[c-\delta, c]$ and $[c, c+ \delta]$ and differentiable at $(c-\delta, c)$ and $(c, c+ \delta)$, for $\epsilon$ small enough (which makes $\delta$ to be small enough and the intervals $[c-\delta, c]$ and $[c, c+ \delta]$ to be contained in $[A,B]$). By the Lagrange's Mean Value Theorem, we have:

$$\exists c_1 \in (c- \delta, c): \ f'(c_1)[c-(c-\delta)]=\delta f'(c_1)=f(c)-f(c - \delta)$$ $$\exists c_2 \in (c, c+ \delta): \ f'(c_2)[(c+\delta)-c]=\delta f'(c_2)=f(c + \delta)- f(c)$$

By uniform continuity of $f$, $\epsilon \rightarrow 0 \implies \delta \rightarrow 0$ and, as $c\in (a,b)$ is fixed, $c_1=c_1(\delta(\epsilon), f), c_2=c_2(\delta(\epsilon),f )$, with:

$$\lim_{\epsilon \rightarrow 0} c_1(\delta(\epsilon), f)= c=\lim_{\epsilon \rightarrow 0} c_2(\delta(\epsilon),f)$$

Then, we finally have:

$$L=\lim_{x \rightarrow c^{-}} f'(x)=\lim_{\epsilon \rightarrow 0} f'(c_1(\delta(\epsilon), f))=\lim_{\epsilon \rightarrow 0} \frac{f(c) - f(c-\delta(\epsilon))}{\delta}=\lim_{x \rightarrow c^{-}} \frac{f(c) - f(x)}{c-x}$$

$$L=\lim_{x \rightarrow c^{+}} f'(x)=\lim_{\epsilon \rightarrow 0} f'(c_2(\delta(\epsilon), f))=\lim_{\epsilon \rightarrow 0} \frac{f(c+\delta(\epsilon))-f(c)}{\delta}=\lim_{x \rightarrow c^{-}} \frac{f(x) - f(c)}{x-c}$$

Since both lateral limits exist and are equal, the theorem is proved and:

$$L=\lim_{x \rightarrow c} \frac{f(x)-f(c)}{x-c}\equiv f'(c)$$

q.e.d."

Am I missing anything? Or is the theorem correctly proved?

• The idea that $c_1,c_2$ tend to $c$ is the key, but you have done it in an unnecessarily complicated manner. Just note that $(f(x) - f(c)) /(x-c) =f'(\xi)$ and as $x\to c$ we have $\xi\to c$ and thus on taking limit we get $f'(c) =L$. May 21, 2018 at 5:07
• Jun 5, 2018 at 3:08

Your argument is on the right track, but it is still obscure. That part you put $\, c_1 \,$ and $\, c_2 \,$ as a function of $\, \delta \,$ and the passage to the limit are not clear. Maybe you can clarify these steps and improve your argument, turning it into a legitimate proof of the corollary.

Allow me show an alternative proof of this result, following the comment by Paramanand Singh. I left a few steps of the proof for you to fill out.

We will use this theorem (which admits the axiom of countable choice) for our purpose:

Theorem of limit by sequences: Let $\ A \subset \mathbb{R} \,$, $\, p \in \mathbb{R} \$ be a limit point of $A$ and $\ \varphi : A \to \mathbb{R} \$ be a function. The limit $\ \displaystyle \lim_{x \to p} \varphi (x) = \ell \$ exists if, and only if, $\ \forall (y_n)_{n \in \mathbb{N}^*} \in (A \setminus \{ p \} )^{\mathbb{N}^*}$, $$\lim_{n \to \infty} y_n = p \ \ \Rightarrow \ \ \lim_{n \to \infty} \varphi (y_n) = \ell \ \ .$$

We want to prove that there exists the limit $\ \, \displaystyle \lim_{h \to 0} \frac{f(c+h)-f(c)}{h} = L \ \,$. Therefore let $\ Q: \ ]a-c,b-c[ \, \setminus \{ 0 \} \to \mathbb{R} \$ be the Newton's quotient function of $f$ at $c$, that is, $\ Q(h) = \frac{f(c+h)-f(c)}{h} \,$, $\forall h \in \ ]a-c,b-c[ \, \setminus \{ 0 \} \$. So we want to prove that $\ \displaystyle \lim_{h \to 0} Q(h) = L \$ and we will do it, using the theorem above.

Proof of the corollary:

Let $\ (s_n)_{n \in \mathbb{N}^*} \in ( \ ]a-c,b-c[ \, \setminus \{ 0 \} )^{\mathbb{N}^*} \$ be such that $\ \displaystyle \lim_{n \to \infty} s_n = 0 \$. I will let you to show that $\ (c + s_n)_{n \in \mathbb{N}^*} \in ( \ ]a,b[ \, \setminus \{ c \} )^{\mathbb{N}^*} \$ and that $\ \displaystyle \lim_{n \to \infty} (c+ s_n) = c \$. Let $\ (J_n)_{n \in \mathbb{N}^*} : \mathbb{N}^* \to \wp( \mathbb{R} ) \$ be the sequence of open intervals given by, $\forall n \in \mathbb{N}^*$, $$J_n = \left\{ \begin{array}{l,l} \ ]c, c+s_n[ \ \ , & s_n >0 \ ; \\ \ ]c+ s_n , c[ \ \ , & s_n < 0 \ . \end{array} \right.$$ Then $\ \big( \overline{J_n} \big)_{n \in \mathbb{N}^*} \in [\wp( \mathbb{R} )]^{\mathbb{N}^*} \$ is the sequence of closed intervals (topological closures) given by, $\forall n \in \mathbb{N}^*$, $$\overline{J_n} = \left\{ \begin{array}{l,l} [c, c+s_n] \ , & s_n >0 \ ; \\ [c+ s_n , c] \ , & s_n < 0 \ . \end{array} \right.$$ It is straightforward that $\ J_n \subset \overline{J_n} \subset \ ] a,b [ \$, $\forall n \in \mathbb{N}^*$.

Let $\ n \in \mathbb{N}^* \,$. By hypothesis (and heredity) $f$ is differentiable on $J_n$ and continuous on $\overline{J_n} \,$. By Lagrange's mean value theorem there exists $\ c_n \in J_n \$ such that $$f'(c_n) = \frac{f(c+ s_n) - f(c)}{s_n} = Q(s_n) \ \ \ .$$ By the theorem of definition by recursion on $\mathbb{N}^*$, there exists a sequence $\ (c_n)_{n \in \mathbb{N}^*} \in ( \ ]a,b[ \, \setminus \{ c \} )^{\mathbb{N}^*} \$ such that $\ c(n) = c_n \$ is given as above, $\forall n \in \mathbb{N}^*$.

Now I ask you to show that $\ \displaystyle \lim_{n \to \infty} c_n = c \$. Hint: for all $\ n \in \mathbb{N}^*$, show that $\ 0 < |c_n - c| < |s_n| \$ and then analyze the sequence $\, (|c_n - c|)_{n \in \mathbb{N}^*} \,$ using the squeeze theorem.

By the theorem of limit by sequences, we have that $$\lim_{n \to \infty} Q(s_n) = \lim_{n \to \infty} f'(c_n) = L \ \ .$$ As the sequence $\, (s_n)_{n \in \mathbb{N}^*} \,$ was chosen arbitrarily, again invoking the theorem of limit by sequences, we have that $\ \displaystyle \lim_{h \to 0} Q(h) = L \$, quod erat demonstrandum. $\blacksquare$

EDIT: I made a mistake in my answer. I wrote

"By the theorem of definition by recursion on $\mathbb{N}^*$, there exists a sequence $\ (c_n)_{n \in \mathbb{N}^*} \in ( \ ]a,b[ \, \setminus \{ c \} )^{\mathbb{N}^*} \$ such that $\ c(n)=c_n \$ is given as above, $\forall n \in \mathbb{N}^*$."

And that's not how you get the sequence. Instead, what we should do is to define a sequence $\ P : \mathbb{N}^* \to \wp( \mathbb{R} ) \$ such that $\ P(n) = P_n = \big\{ r \in J_n : f'(r) = Q(s_n) \big\} \,$. Then, by Lagrange's mean value theorem, we have that $\ P_n \neq \varnothing \,$, $\forall n \in \mathbb{N}^*$. So the set $\ im(P) = \{ P_n \in \wp( \mathbb{R} ) : n \in \mathbb{N}^* \} \$ is a countable non-empty set whose elements are non-empty sets. By the axiom of countable choice there exists a choice function for $\, im(P) \,$, that is, a function $$\psi : im(P) \to \bigcup im(P) = \bigcup_{n \in \mathbb{N}^*} P_n$$ such that $\ \psi(R) \in R \,$, $\forall R \in im(P)$. In other words $\ \psi(P_n) \in P_n \,$, $\forall n \in \mathbb{N}^*$. Thus the function $\ \psi \circ P \$ is precisely the sequence $\, (c_n)_{n \in \mathbb{N}^*} \,$ we are looking for.