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My textbook defines $p = f'(a^{+})$ and $q= f'({a^-})$, and says;

(I) p=q =>f is differentiable at x = a => f is continuous at x = a

(II)p$\ne$q => f is not differentiable at x = a, but f is continuous at x=a.

How does the second one work out? It needn't be continuous, right?

harry
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  • One thing to take note of is $p$ and $q$ are differentials. So at $a$, $f$ have to be continuous to allow $p$ and $q$ to be defined. – Michael Teguh Laksana Nov 15 '20 at 10:38
  • Aren't p and q differentials of f(x) at a+h and a-h, and not at a? Like, for f(x)={|x|, when x$\ne$0, 1, when x=0} f'(a+h) and f'(a-h) are defined, but the function isn't continuous at x=0. – harry Nov 15 '20 at 12:18
  • Not exactly. A differential of $f$ at point $a$ ($f'(a)$) is defined as the gradient between $f(a)$ and $f(a+h)$ where $\lim_{h\rightarrow a}$. So $p$ and $q$ are both differentials at point $a$, but from different directions. Another thing is that for $f(x)=|x|$, x is continuous at x=0! (Continuous does not imply smooth! Continuity is defined by the limit from both left and right direction to be equal). It is however not differentiable (aka not smooth). – Michael Teguh Laksana Nov 15 '20 at 12:28
  • Oh, so $p = \frac{f(a+h)-f(a)}{h}$ and $q = \frac{f(a)-f(a-h)}{h}$, when h is a miniscule positive value, right? But why exactly does f have to be continuous for either of these to exist? And also, I didn't say |x|, i said f(x)=0 at x=0, and f(x)=|x| everywhere else. – harry Nov 15 '20 at 12:40
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    $p$ and $q$ can exist without continuity of $f$ at $a$. Keep in mind $p$ and $q$ are both either from left or right limit of $f$. What would not exist if $f$ is discontinuous at $a$ is $f'(a)$, as can be seen from the inequality of $p$ and $q$. – Michael Teguh Laksana Nov 15 '20 at 12:49

1 Answers1

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If $f(x)$ is defined in some neighbourhood of $a\in\mathbb R$ and both $\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}$ and $\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$ exist (as finite numbers), then $f(x)$ will be continuous at $x=a$.

This is because $f(x)=f(a)+(x-a)\frac{f(x)-f(a)}{x-a}$ so, if, say, $\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}=p$, then $\lim_{x\to a^-}f(x)=f(a)+0\cdot p= f(a)$, and similar for the limit "from above".

harry
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  • What if there's a jump discontinuity at a? Then your conditions would be satisfied and f would still be discontinuous. – harry Nov 15 '20 at 12:44
  • @HarryHolmes You did not say how you've defined $f'(a^{-})$ or $f'(a^{+})$. If they are defined as I define them above, then on the jump discontinuity at least one of them, or maybe both, won't exist. Note the definitions above explicitly involve $f(a)$.If by $f'(a^{-})$ you mean $\lim_{x\to a^{-}}f'(x)$, then this is a different beast and obeys different laws. –  Nov 15 '20 at 14:15
  • They're not defined in any particular way; the textbook states it as a general rule. Wouldn't it go wrong if there was a jump discontinuity? As in, $f(x)= x+3$ if $x \geq 0$, and $-3-x$ for $x \lt 0$? – harry Nov 16 '20 at 01:40
  • @HarryHolmes If $f(x)=x+3$ for $x\ge 0$, it means $f(0)=3$ and thus $\lim_{x\to 0^{-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0^{-}}\frac{-6-x}{x}=\infty$ so $f'(0^{-})$ does not exist. Compare that with $\lim_{x\to 0^{-}}f'(x)=\lim_{x\to 0^{-}}-1 = -1$ which does exist. My point is - both $\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$ and $\lim_{x\to a^{-}}f'(x)$ tell you something about the behaviour of the function on the "left" side of $a$ - but they tell you different things. My guess is that your textbook either explicitly or implicitly adopts the first formula as the definition of $f'(a^{-})$. –  Nov 16 '20 at 09:51
  • Yes, aren't they the same? – harry Nov 17 '20 at 03:18
  • "Aren't they the same" - who are "they"? Anyways, I am glad you have provided this link - they use the same definition of the left and right derivative that I've been using. With this definition, as I said, if any of those one-sided derivatives exists (as a finite number), then the function must be at least one-sided continuous in $a$ (on that side). I am struggling to understand if there are any further points that you want to have clarified. –  Nov 17 '20 at 11:03
  • By "they", I meant $\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$ and $\lim_{x\to a^{-}}f'(x)$, that you said weren't equal yesterday. Just this; aren't $f'{-}(a)$ and $\lim{x\to a^{-}}f'(x)$ equal? Because the answer I linked says that $f'{-}(a)$ is equal to $\lim{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$. If they are equal, it would imply that your earlier statement was incorrect. – harry Nov 17 '20 at 12:30
  • I maintain that $f'{-}(a)$ and $\lim{x\to a^{-}}f'(x)$ are not the same thing - which your sample above demonstrates. ($f'{-}(0)$ does not exist, while $\lim{x\to 0^{-}}f'(x)=-1$ exists). I am not saying those two limits are not related: in fact, if the function happens to be left-continuous when $x\to a^{-}$ and if $\lim_{x\to a^{-}}f'(x)$ exists, then $f'_{-}(a)$ also exists and is equal to the previous (you can use mean value theorem to prove this!) but you have to assume (left-)continuity - cannot prove it. –  Nov 17 '20 at 14:22
  • (Cont'd): The answer you linked does not say anything similar to what you claim it says, so I am not sure I fully understand what you mean. It only says that, if $f'{-}(a)$ and $f'{+}(a)$ exist (i.e. the left- and right-side derivative exist) and are equal, then the "full" both-sided derivative $f'(a)$ exists and is equal to both. –  Nov 17 '20 at 14:25
  • (Cont'd): Note $f'{-}(a)$ may exist without $\lim{x\to a^{-}}f'(x)$ existing too, in fact the function doesn't need to be differentiable (or even continuous) in any left neighbourhood of $a$. Look at the function $f(x)=\begin{cases}x^2&x\in\mathbb Q\0&x\not\in\mathbb Q\end{cases}$. This function is continuous and differentiable at $x=0$ and is not even continuous (let alone differentiable) at any other point. –  Nov 17 '20 at 14:28
  • (Cont'd): I am guessing that you may be confused about the definition of $f'{-}(a)$. As the other post says, it is defined as $\lim{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$. It is not defined as $\lim_{x\to a^{-}}f'(x)$ and nobody claims that that limit is equal to $\lim_{x\to a^{-}}\frac{f(x)-f(a)}{x-a}$. –  Nov 17 '20 at 16:24
  • Yeah, I'm confused about the notation. Since this thread got a bit messy, I posted a question, and it looks like all the notations mean the same thing. – harry Nov 17 '20 at 17:38
  • Yep, I've been trying to say that third and fourth are not the same. Cf. the answer (the only one you've got by now): https://math.stackexchange.com/a/3911571/700480 –  Nov 17 '20 at 20:33