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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)?

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    $\begingroup$ If by "defined" you mean being finite, then yes, it does. $\endgroup$ Aug 9 at 14:25
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    $\begingroup$ Let's consider $$f(x)=\begin{cases}x, &x>0\\ -x-1,& x<0 \end{cases}$$ what do you think about? $\endgroup$
    – zkutch
    Aug 9 at 14:35
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    $\begingroup$ We have $$ \lim_{h\rightarrow 0^\pm} (f(a+h)-f(a))=\lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} h = \lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} \cdot \lim_{h\rightarrow 0^\pm} h =\lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} \cdot 0= 0.$$ $\endgroup$ Aug 9 at 14:42
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    $\begingroup$ @SeverinSchraven Thank you very much. As Sourav said, would you mind making it an answer? $\endgroup$
    – MangoPizza
    Aug 9 at 15:00
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    $\begingroup$ @zkutch: In the question as posted, $f(a)$ appears as a term under the limit sign. So $f(a)$ has to be assigned a value in order for the post to make sense. But your example does not assign a value to $f(0)$, nor can a value even be assigned which makes both limits in the post exist as required. $\endgroup$
    – Lee Mosher
    Aug 9 at 15:43

3 Answers 3

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$\lim_{x\to a}f(x) =L$ iff $\lim _{x→a ^{−}} f (x) = L = \lim_{ x→a^{ +}} f (x)$ $\space$$($see here $)$


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

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  • $\begingroup$ Left hand derivative is not same with derivative from left. Source in my last comment. $\endgroup$
    – zkutch
    Aug 9 at 15:41
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    $\begingroup$ @zkutch How would you define "left hand derivative " and "derivative from the left"? proofwiki.org/wiki/…. $\endgroup$ Aug 9 at 15:44
  • $\begingroup$ Answer is in Olmsted's book mentioned above. $\endgroup$
    – zkutch
    Aug 9 at 15:47
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    $\begingroup$ I like this answer :) $\endgroup$ Aug 10 at 0:15
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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.

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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 $a$.

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