Is it true that

If $f(x)$ is differentiable at $a$, then both $f'(a^+)$ and $f'(a^-)$ exist and $f'(a^+)=f'(a^-)=f'(a)$.

My answer is NO.

Consider the function $$ f(x)=\begin{cases} x^2\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex] 0&\text{for $x=0$} \end{cases} $$

$f'(0)$ can be found by

\begin{align} \lim_{x \to 0} \dfrac{f(x) - f(0)}{x-0} & = \lim_{x \to 0} \dfrac{f(x) - 0}{x} & \textrm{ as } f(0) = 0 \\ & = \lim_{x \to 0} \dfrac{x^2 \sin\left(\frac{1}{x}\right)}{x} & \\ & = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) & \end{align}

Now we can use the Squeeze Theorem. As $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we have that $$0 = \lim_{x \to 0} x \cdot -1 \leq \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \leq \lim_{x \to 0} x \cdot 1 = 0$$

Therefore, $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$ and we have $f'(0)=0$.

However, $$ f'(x)=\begin{cases} -\cos\dfrac{1}{x}+2x\sin\dfrac{1}{x}&\text{for $x\ne0$}\\[1ex] 0&\text{for $x=0$} \end{cases} $$ $f'(0^+)$ nor $f'(0^-)$ exists as $x\to 0$.

Is my answer correct?

I have found some pages related to this question.

Is $f'$ continuous at $0$ if $f(x)=x^2\sin(1/x)$

Calculating derivative by definition vs not by definition

Differentiability of $f(x) = x^2 \sin{\frac{1}{x}}$ and $f'$

$f'$ exists, but $\lim \frac{f(x)-f(y)}{x-y}$ does not exist


  • 2
    $\begingroup$ I'm not familiar with the notation. Does $f'(a^+)$ mean $\lim_{x \to a^+} f'(x)$? $\endgroup$ – pjs36 Jan 14 '17 at 5:21
  • $\begingroup$ Your example works. $\endgroup$ – user384138 Jan 14 '17 at 5:22
  • 1
    $\begingroup$ @Wong Austin But the right derivative of $f$ at $a$ is given by$$\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$$ $\endgroup$ – Juniven Jan 14 '17 at 5:30
  • $\begingroup$ Supposing that it ($f'(a^{+})$) means right derivative makes your question trivial because derivative at a point exists if and only both left and right derivatives at the point exist and are equal. The example you have given in your post uses a different meaning of the symbol $f'(a^{+})$ which matches the comment from @pjs36 as well as my answer. $\endgroup$ – Paramanand Singh Jan 14 '17 at 5:45

Your working is correct. I assume you mean $$f'(a^{+}) = \lim_{x \to a^{+}}f'(x), f'(a^{-}) = \lim_{x \to a^{-}}f'(x)$$ Just to clarify notation, $f'(a^{+})$ does not mean right hand derivative of $f$ at $a$ but rather it means right hand limit of the derived function $f'$. Similar remark applies to $f'(a^{-})$. I believe OP is using these symbols in the manner I have explained above.

Existence of a derivative at a point does not necessarily mean that it is continuous at that point. The example you have given in your post is a classic example of such a scenario when $f'$ exists but is not continuous.

What is important is to know that a derivative can not have jump discontinuity. Thus if $f'(a^{+}), f'(a^{-})$ exist and $f$ is continuous at $a$ then $f'(a)$ also exists and $f'(a) = f'(a^{+}) = f'(a^{-})$. Note that in your example $f'(a)$ exists but both the limits $f'(a^{+}), f'(a{-})$ do not exist. Had they existed $f'$ would have been continuous at $a$.

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  • $\begingroup$ Thank you. I was confused $f'(a^+)$ with the right hand derivative:) $\endgroup$ – Tianlalu Jan 14 '17 at 6:15

$$\lim_{x \to x_0^+} f'(x)$$

is not the same as

$$\lim_{x \to x_0^+} \frac{f(x) - f(x_0)}{x-x_0}$$

The latter limit always exists if $f'$ does, and it is usually what we mean when we say "right derivative".

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Yes, your answer is correct. The existence of the derivative of a function at a point does not always mean that the derivative will be continuous at that point. The condition $f′(a+)=f′(a−)=f′(a)$ implies continuity of the derivative at $x=a$ which is clearly not true for the function you mentioned at $x=0$.

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