Does Left Hand Derivate and Right Hand Derivative being defined guarantee continuity? Suppose at $x = a$, both the Left Hand Derivative and Right Hand Derivative of a function exists and is defined. In other words, both the limits $$\lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}$$ and $$\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}$$ exist and are defined.
Does that guarantee that the function is continuous at $x = a$ (no matter the limits are equal or not)?
 A: $\lim_{x\to a}f(x) =L$ iff $\lim _{x→a ^{−}} f (x) = L = \lim_{ x→a^{ +}} f (x)$
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If $\lim_{x\to a}f(x) =f(a) $ i.e $\lim_{h\to 0} f(a+h)=f(a)$ then $f$ is continuous at $a$.

Let $\lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}=l$
$\begin{align}\lim_{h \to 0^-} f(a+h) -f(a) &= \lim_{h \to 0^-}  \frac{f(a+h)-f(a)}{h}\cdot h\\&= \lim_{h \to 0^-}  \frac{f(a+h)-f(a)}{h} \cdot \lim_{h\to 0^-} h\\&=l\cdot 0\\&=0
 \end{align}$
Similarly  $\lim_{h \to 0^+} \frac{f(a+h)-f(a)}{h}=L$ implies
$\lim_{h \to 0^+} f(a+h) -f(a)=0$

$\lim_{h\to 0^+}f(a+h) =f(a) =\lim_{h\to 0^-}f(a+h) $
Hence $f$ is continuous at $a$.
Credit: Severin Schraven
A: For the derivative to exist at $a$ means there is an open set in the domain that contains $a$ where the derivative will be defined. Now since the function is continuous everywhere it's derivative is defined we have that the function is both left and right continuous at $a$. However since the open sets for the right and left hand derivative both contain $a$ it forces the left and right hand limits to be the same.
A: Let $L^{+}$ denote the right hand limit. For $h > 0$ we have $f(a + h) = f(a) + hL^+ + o(h)$. Clearly $hL^+ + o(h) \to 0$ as $h \to 0$, so $f$ is right-continuous at $a$. The same argument shows that $f$ is left-continuous at $a$. Hence $f$ is continuous at $a$.
