Existence of right-hand and left-hand derivatives at a point implies differentiability at the point?

Question

Consider a real-valued function $f:[a,b]\rightarrow \mathbb{R}$ and consider some point $c\in (a,b)$. Assume that $$f^\prime(c^-):=\lim_{x\nearrow c}\frac{f(x)-f(c)}{x-c}\quad\text{and}\quad f^\prime(c^+):=\lim_{x\searrow c}\frac{f(x)-f(c)}{x-c}$$ both exist. That is, assume that the left-hand and right-hand derivatives of the function exist at the point $c$. Further assume that $f^\prime(c^-)=f^\prime(c^+)$, i.e., that the left-hand and right-hand derivatives at the point $c$ equal. Does this necessarily imply that $f$ is differentiable at $c$, i.e., that $f^\prime(c):=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}$ exists? Could you provide a proof or reference for this fact? If it doesn't necessarily imply differentiability, are there additional conditions that would imply differentiability?

Intuitively, it seems to be true. But I couldn't prove it. I saw a few related posts on Math Stack Exchange (with one directly asking this question), but for some reason I didn't see a convincing enough answer there.

Edit: There is something that I want to add here and ask additionally. Suppose you know the derivative of $f$ to both the left and right of $c$, but not at $c$. Suppose you can show that $$\lim_{x\nearrow c} f^\prime(x)=\lim_{x\searrow c} f^\prime(x).$$ Can it be concluded from this that the derivative of $f$ at $c$, i.e., $f^\prime(c)$, exists and equals those left-hand and right-hand limits? Furthermore, can it be concluded that $f^\prime$ is continuous at $c$?

Answer 1

The derivative of a function $f$ exists if the right-hand and left-hand derivative of $f$ exist, and both of them are equal.

Indeed, let $(x'_n)_{n\in\mathbb{N}}$ be any sequence in $[a, b]$ convergent to $c$ s.t. $x'_n \neq c$ for all $n \in \mathbb{N}$. Take an arbitrary subsequence $(x_n)_{n\in\mathbb{N}}$ of $(x'_n)$.

Then we can take a further subsequence $(z_n)_{n\in\mathbb{N}}$ s.t. $z_n < c$ for all $n \in \mathbb{N}$, or $z_n > c$ for all $n \in \mathbb{N}$ since $(x_n)$ must have infinitely many $n\in \mathbb{N}$ s.t. $x_n < c$ or have infinitely many $n$ s.t. $x_n > c$.

Since both right-side derivative and left-side derivative exist and they are equal, we have $$\lim_{n\to\infty} \frac{f(z_n)-f(c)}{z_n-c} = f'(c^-)=f'(c^+)$$ for any case.

We showed that for any subsequence $(\frac{f(x_n)-f(c)}{x_n-c})_n$ of $(\frac{f(x'_n)-f(c)}{x'_n-c})_n$ we have a further subsequence $(\frac{f(z_n)-f(c)}{z_n-c})$ convergent to $f'(c^-)=f'(c^+)$. Therefore, since if for any subsequence of a sequence there exists a further subsequence converging then a sequence converges, $$\lim_{n\to\infty} \frac{f(x'_n)-f(c)}{x'_n-c} = f'(c^-) = f'(c^+)$$

Thus $f'(c)$ exists and $f'(c)=f'(c^-)=f'(c^+)$

Answer 2

Theorem: A function $f$ is differentiable at $c$ if and only if the left and right limits of its difference quotient exist and are equal, i.e., $ f'(c^-) = f'(c^+) = L $ for some real number $ L $.

Proof:

We’ll prove each direction.

Forward Direction: Assume $ f $ is differentiable at $ c $, which means the limit \begin{equation} f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{equation} exists. By the sequential criterion for limits, this implies that for any sequence $ x_n \to c $, the sequence of difference quotients \begin{equation} \frac{f(x_n) - f(c)}{x_n - c} \to f'(c). \end{equation} In particular, this holds for sequences $ (x_n) $ approaching $ c $ from above or from below. Thus, both the right-hand limit $ f'(c^+) $ and the left-hand limit $ f'(c^-) $ must exist and equal $ f'(c) $. Therefore, we conclude that $ f'(c^-) = f'(c^+) = f'(c) $.

Reverse Direction: Assume that the left and right limits of the difference quotient exist and are equal; let $ f'(c^+) = f'(c^-) = L $. We’ll show that this implies $ f $ is differentiable at $ c $ with $ f'(c) = L $.

Consider any sequence $ x_n \to c $. We can divide $ (x_n) $ into two subsequences: $ (y_m) $ (where $ y_m > c $ for all $ m $) and $ (z_l) $ (where $ z_l < c $ for all $ l $), though it may happen that one of these subsequences is finite or empty. At least one of them, however, must be infinite, and in any of these scenarios we have that $\forall n, x_n=y_m$ or $x_n=z_m$ with $y_m \rightarrow c^+$ and $z_l \rightarrow c^-$. So, we have either: \begin{equation} \frac{f(x_n) - f(c)}{x_n - c} =\frac{f(y_m) - f(c)}{y_m - c} \rightarrow f'(c^+) = L, \end{equation} or \begin{equation} \frac{f(x_n) - f(c)}{x_n - c} =\frac{f(z_l) - f(c)}{z_l - c} \rightarrow f'(c^-) = L. \end{equation} Thus, regardless of the approach direction, the difference quotient converges to $ L $, meaning that \begin{equation} \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = L. \end{equation} This proves $ f $ is differentiable at $ c $ with $ f'(c) = L $.

Of course, the last part should be made more rigerous using the $\epsilon-N$ convergence of the sequences.

Edit

For a function $ g $ to be continuous at $ c $, we must have $ \lim_{x \rightarrow c} g(x) = g(c) $. The argument above holds if we replace $ f'(x) $ by a function $ g(x) $; that is, the limit exists if and only if $ g(c^+) = g(c^-) $, or alternatively, this means a function is continuous if and only if $ g(c^+) = g(c) = g(c^-) $. This holds true in the case of $ f'(x) $ when $ f $ is differentiable.