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From this discussion : Why is the absolute value function not differentiable at $x=0$?

I understand the idea of computing the left (n-) and right (n+) limits and if they are a different number, then we consider that the general limit of of x when x->n does not exist. This was explained for the |x| and it makes sense.

However I tried to do the same for $x^2$ which is given as an example of a differentiable function (especially in the point where x=0), and i end up to 2 different left and right limits which would mean that $x^2$ would not be differentiable in 0 .... which I know (or at least i think i know? is false - but not so sure now...).

this is how i did, using the derivative's "limit definition" as in the link provided

I imagine that i made a mistake of course... but i really don't see where, as I strictly repeated the same procedure as for |x|. Could someone explain if possible in detail why?

SheppLogan
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3 Answers3

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$0^+$ and $0^-$ are considered to be equal for the purposes of deciding whether the limits from $+$ and $-$ are the "same" value. Note that when you try the same thing with $|x$ you get $+1$ and $-1$ and these are of course not equal.

Mark Fischler
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  • Just to add to this, it has to do with how the real numbers are defined: on one definition, they are "equivalence classes of Cauchy sequences of rational numbers," so $-1+0.999\ldots=1-0.999\ldots$, which you might think of as $0^{-}=0^{+}.$ – RideTheWavelet Jun 28 '17 at 23:34
  • really 0+ and 0- equal ? could you detail more on why they are equal ? – SheppLogan Jun 28 '17 at 23:35
  • it seems very unintuitive to say that a negative number be equal to a positive one – SheppLogan Jun 28 '17 at 23:36
  • but thanks for the very quick answers by the way ;) – SheppLogan Jun 28 '17 at 23:37
  • I ve seen proofs for the fact that 0.999... =1 so i can admit that, yet that does not help me to understand why 0- = 0+ by the argument provided by RideTheWavelet – SheppLogan Jun 28 '17 at 23:39
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    @Mark Fischler I am not aware of any numbers $0^+$ or $0^-$. Why not just say $\lim_{h \to 0^+} h = 0$? – Ovi Jun 28 '17 at 23:57
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I think we need to take this back to what is the definition of a limit.

$\lim_\limits{h\to 0} \frac {h^2}{h} = 0$

means

For any $\epsilon>0$ there exists a $\delta > 0$ such that $| \frac {h^2}{h}|<\epsilon$ when $|h|<\delta$

For every neighborhood around $0$ in the image of $f(h) = \frac {h^2}{h}$ (sometimes called an epsilon ball) that we define, we can find an a neighborhood around $0$ in the domain (called the delta ball) such that everything in the delta ball maps inside the epsilon ball.

This is not a particularly easy concept, and may take some time to "grok in fullness."

left hand and right hand limits.

$\lim_\limits{x\to 0^+} \frac {h^2}{h} = 0$ means

$\forall \epsilon>0,\exists \delta >0: 0<h<\delta \implies |\frac {h^2}{h}-0|<\epsilon$

and $\lim_\limits{h\to 0^-} \frac {h^2}{h} = 0$ means

$\forall \epsilon>0,\exists \delta >0: 0>h>-\delta \implies |\frac {h^2}{h}-0|<\epsilon$

These limits do not equal $0^+$ and $0^-$ respectively, they equal exactly $0.$

Doug M
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  • Thanks for the rigourous answer. Indeed, a bit difficult to understand quickly because of the many nested concepts needed to be understood simultaneously... but that's often my problem... – SheppLogan Jun 29 '17 at 00:09
  • The short answer $0^+$ and $0^-$ are not real numbers. $0$ is a real number. $0^+$ notation says something about the neigborhoods you are examining. But the value of the limit is a real number -- if it exists, and assuming that you are working over the real numbers. I suppose the limit can be $-\infty$ or $\infty$ So, more precisely, the limit will be in the extended real numbers. – Doug M Jun 29 '17 at 00:17
  • ok but it is admited that 0- is negative ? I seems strange that in the case of |x| the 0- puts the thing to negative – SheppLogan Jun 29 '17 at 00:19
  • Most calculus students do not get a grounding in the formal definitions of limits. It is put up on the board, and then quickly forgotten. Those who stick with math, then cover it in depth in Real Analysis class. – Doug M Jun 29 '17 at 00:19
  • yes unfortunately i an a computer scientist (but i still do quite a bit of maths) and i am interested in maths. But not a mathematician ! – SheppLogan Jun 29 '17 at 00:20
  • No, I do not concede that $0^-$ is negative. It is not a real number, and cannot be ordered with the real numbers. It is a set. – Doug M Jun 29 '17 at 00:22
  • humm ok. I think i was confused by something in here https://math.stackexchange.com/questions/991475/why-is-the-absolute-value-function-not-differentiable-at-x-0/991559 but indeed it was the definition of the abs that put the left limit to negative and not 0-, just by algebraic simplification of -h/h=-1 – SheppLogan Jun 29 '17 at 00:30
  • That question says $\lim_\limits{x\to 0^+} \frac {|h|}{h} = 1, \lim_\limits{x\to 0^-} \frac {|h|}{h} = -1$ and $\lim_\limits{x\to 0^+} \frac {h^2}{h} = \lim_\limits{x\to 0^-} \frac {h^2}{h} =\lim_\limits{x\to 0} \frac {h^2}{h} =0$ – Doug M Jun 29 '17 at 00:36
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There is no number called "$0^+$".

And there is no number called "$0^-$".

There is only a number called "$0$".

When you write $$\lim_{h \to 0^+} h = 0^+ $$ it does not mean that the limit is equal to "the number $0^+$" because, as just explained, there is no number called "$0^+$".

Instead, what it means is that as $h$ approaches $0$ from above, the limit of $h$ is equal to the number $0$ and, furthermore, it approaches that number from above.

Lee Mosher
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