# 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)?

• If by "defined" you mean being finite, then yes, it does. Aug 9 at 14:25
• Let's consider $$f(x)=\begin{cases}x, &x>0\\ -x-1,& x<0 \end{cases}$$ what do you think about? Aug 9 at 14:35
• We have $$\lim_{h\rightarrow 0^\pm} (f(a+h)-f(a))=\lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} h = \lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} \cdot \lim_{h\rightarrow 0^\pm} h =\lim_{h\rightarrow 0^\pm} \frac{f(a+h)-f(a)}{h} \cdot 0= 0.$$ Aug 9 at 14:42
• @SeverinSchraven Thank you very much. As Sourav said, would you mind making it an answer? Aug 9 at 15:00
• @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. Aug 9 at 15:43

$$\lim_{x\to a}f(x) =L$$ 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

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

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.

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 $$a$$.