I am reading a definition which claims that a function is continuous in point $p$ iff all its first derivations exist and are continuous in the point $p$.

And what confuses me are functions such as $f(x)=|x|$ which should be continuous by intuition, but is clearly not derivable in $x=0$.

I am almost certain I am getting something wrong here, but I can not even pin-point what.

  • $\begingroup$ For $|x|$ its derivative isn't continuous t zero. $\endgroup$
    – coffeemath
    Apr 9, 2019 at 20:51
  • $\begingroup$ Where did you read that erroneous definition? $\endgroup$
    – bof
    Apr 9, 2019 at 21:01
  • $\begingroup$ lecture notes by my prof. i might be mosreading them though $\endgroup$
    – fazan
    Apr 9, 2019 at 21:04
  • 1
    $\begingroup$ @avs That is false. $\endgroup$
    – zhw.
    Apr 9, 2019 at 22:07
  • 2
    $\begingroup$ @avs That is the definition of a continuously differentiable or $C^1$ function. Being differentiable is strictly weaker (not requiring that the derivatives be continuous). $\endgroup$ Apr 10, 2019 at 0:19

3 Answers 3


That "definition" is wrong. You are right, the function $|x|$ is continuous but is not differentiable at $x=0$. Continuity doesn't imply differentiability. However, differentiability does imply continuity.

The definition you stated looks to me as an attempt to define a smooth function, although it is not correct.

  • $\begingroup$ It is the definition of a continuously differentiable or $C^1$ function. This definition is important because $C^1$ functions on compact manifolds form Banach spaces, whereas differentiable functions do not. $\endgroup$ Apr 10, 2019 at 0:20
  • $\begingroup$ Here is the relevant wikipedia page: en.wikipedia.org/wiki/… $\endgroup$ Apr 10, 2019 at 0:24

As has been pointed out this definition is incorrect, as it is inconsistent with the usual definitions of continuity and differentiability. Your example $|x|$ suffices to show this.

If you are encountering this in multivariable calculus then your professor might be trying to state the theorem mentioned by avs in the comments: that a function is differentiable at a point if all its first order partial derivatives exist in a neighbourhood of that point, and are continuous at that point. However the converse is not generally true: consider for example the function $$f(x,y)=\begin{cases}(x^2+y^2)\sin(\frac{1}{\sqrt{x^2+y^2}}) &(x,y)\neq(0,0)\\0&(x,y)=(0,0)\end{cases}$$ at the origin. Thus this assumption might be completely false. It might be best to give a word for word reproduction of the statement and the paragraph before and after.

  • $\begingroup$ The partial derivatives need not be continuous for differentiability. $\endgroup$ Apr 9, 2019 at 21:13
  • 1
    $\begingroup$ @HarisGusic yes I realized as I posted. Fixed it $\endgroup$
    – K.Power
    Apr 9, 2019 at 21:13

Sounds like somebody got the wrong definition of what a "continuous function" is. Any function $f:\mathbb{R}\to\mathbb {R}$ (like in your original post!) is continuous at any point $\left(a,f\left(a\right)\right)$ for which $$\lim\limits_{x\to a^-}f\left(x\right)=\lim\limits_{x\to a^+}f\left(x\right)$$ (denoting the left and right-hand limits accordingly and provided both limits exist).

And finally, note that some functions can even be nowhere-continuous as well! Such as $$f\left(x\right)=\left\{\begin{matrix}1, x\in\mathbb{Q}\\0,x\notin\mathbb {Q}\end{matrix}\right.$$


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