17
$\begingroup$

I don't really understand why a discontinuous function cannot be differentiable.

In Stewart's Calculus, the definition of a function $f$ being differentiable at $a$ is that $f'(a)$ exists. Earlier it gives the definition of the derivative as $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. It also has a theorem that if $f$ is differentiable at $a$, then $f$ is continuous at $a$. There is also a definition that a functions $f$ is continuous at a number $a$ if $\lim_{x\to a}f(x)=f(a)$.

Take for example the very simple function:

$$ f(x)=\begin{cases} x+1 & x\geq0, \\ x & x<0. \end{cases} $$

It is discontinuous at $x=0$ (the limit $\lim_{x\to 0}f(x)$ does not exist and so does not equal $f(0)$), but if I find the derivative using the limit above, I get the left and right limits to equal $1$. So therefore, the derivative exists.

According to the book, the function shouldn't be differentiable at $x=0$ as it has a discontinuity (continuity is a necessary condition of differentiability). What am I doing/understanding wrong?

$\endgroup$
4
  • 6
    $\begingroup$ The limit from the left would give you $$\lim\limits_{h\rightarrow0^-}{f(0+h)-f(0)\over h } =\lim\limits_{h\rightarrow0^-}{h -1\over h } =\infty.$$ $\endgroup$ Commented Jun 9, 2014 at 22:49
  • $\begingroup$ Thank you @DavidMitra. I completely missed that. I don't think I can vote that as the answer since it's a comment. $\endgroup$ Commented Jun 9, 2014 at 23:11
  • 10
    $\begingroup$ On a rough intuitive level, continuous means "looks connected as you zoom in", and differentiable means "looks like a line segment as you zoom in". It can't look like a line segment without looking connected. $\endgroup$ Commented Mar 18, 2016 at 12:44
  • 1
    $\begingroup$ @Mark. Except in the case of vertical tangent. $\endgroup$ Commented Aug 19, 2022 at 7:58

2 Answers 2

Reset to default
10
$\begingroup$

Computing the derivative from the left gives you $$ \lim\limits_{h\rightarrow0^-} {f(0+h)-f(0)\over h } =\lim\limits_{h\rightarrow0^-} {h-1\over h }=\infty. $$ (In particular, note $f(0)=1$, not $0$.)

You can also see the derivative from the left doesn't exist (as a real number) by considering slopes of secant lines. Note a secant line has one endpoint at the point $(0,1)$ and the other at a point $(h,h)$ with $h<0$. As $h$ tends to $0$, the slopes tend to $\infty$.

$\endgroup$
3
$\begingroup$

You are computing $\lim_{x\rightarrow 0_{\pm}}f'(x)$ rather than $$\lim_{h\rightarrow 0^\pm}\frac{f(0+h)-f(0)}{h}.$$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.