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I recently saw this article by oregonstate.edu about the differentiability of the function $y=\sqrt{X}$ at $x=0$ they claim that the derivative at $x=0$ does not exist.i understand the concept of one-sided differentiability and how the $\lim_{h \to 0-} \frac{\sqrt{x+h}-\sqrt{x}}{h} $ does not exist but for values of greater than 0 and approaching from the right they say that the limit does not exist as it is positive infinity. my question is why isn't this limit defined surely we know it is approaching infinity so why can't we say it has an infinite slope

tryst with freedom
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The homeschooler
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    It's a matter of definition. The problem is obviously that the function is not defined for any negative value. You cannot compute the derivative using the definition as a limit of the difference quotient since the limit does not exist because the function is not defined in any left neighborhood of 0. Then of course $\lim_{h \to 0+} \frac{1} {\sqrt h} =+\infty $. – Adriano Banchieri Dec 10 '20 at 09:23
  • Anyway, for such reasons, infinite slopes are usually referred to only when both left and right derivatives exist and they are $+$ or $-\infty $. – Adriano Banchieri Dec 10 '20 at 09:29
  • You don't say "is infinite". You say "does not exist". –  Dec 10 '20 at 09:39
  • I think you should make the title more general @The homeschooler – tryst with freedom Dec 10 '20 at 09:41
  • The limit from the right is positive infinity – The homeschooler Dec 10 '20 at 09:41
  • It is typical in beginning calculus to not deal with infinite derivatives, and even in more advanced mathematics in which they might be discussed, it is typical for "differentiable" to mean "finitely differentiable". One can certainly discuss infinite derivatives (see here and here), whether two-sided signed versions, two-sided unsigned versions, one-sided signed versions, or one-sided unsigned versions ("signed" means the difference quotient limit (continued) – Dave L. Renfro Dec 10 '20 at 10:49
  • must approach one of $-\infty$ and $+\infty,$ and "unsigned means the absolute value of the difference quotient limit approaches $\infty).$ In the case of $\sqrt{x},$ there is a right-sided derivative of $+\infty$ at $x=0.$ Usually when one wants to talk about allowing for infinite derivatives, one says things like "the derivative exists, finitely or infinitely" (and the "signed" version is nearly always intended when nothing explicit is said about signs). – Dave L. Renfro Dec 10 '20 at 10:54
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    Why is "this article" not a link to it? I'd like to read what exactly it says, and that could explain it. – superb rain Dec 10 '20 at 17:34
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    @superb rain: I too wonder why links, specific cites, etc. are often not included in certain questions (and I'm talking about those it's clear the OP knows them, rather than cases when the OP is trying to conceal the asking of a homework question). To me it seems obvious one would want to provide this information to others, and it's not an age or maturity issue I think, as I feel sure I would have done so even in high school (if internet math groups had existed and been available to me). Indeed, at my high school when I attended, NO ONE (not even myself, (continued) – Dave L. Renfro Dec 10 '20 at 19:11
  • even though I was beginning to study calculus (from a library book) before finishing 9th grade) took Algebra 1 before 9th grade (i.e. in Junior High), but we were still thoroughly taught about citing, as appropriate, specific textbook pages (or in library magazines or library books) and such in history and other assignments when answering essay questions for homework (or essays written when the class went to the library to allow us to research a topic). – Dave L. Renfro Dec 10 '20 at 19:16
  • @DaveL.Renfro Yes, they really should want to include it, given that we could then tell them exactly what/how they misunderstood (or in case it's in fact wrong, we could then confirm that). Apparently they're not interested in that. – superb rain Dec 10 '20 at 19:22
  • I'm sorry but im not quite sure how to share a link on here – The homeschooler Dec 11 '20 at 12:02
  • For sharing a link see this. FYI, I didn't know for the first year or two I was here, and I simply gave links the same way everyone else had done so for well over 10 years before Math. Stack Exchange began (for me, mostly in sci.math), which was to simply paste the URL into the message (as a separate line, if line breaks were respected) so that the reader could simply copy the URL and paste it into an internet browser's "URL location bar". Finally, someone told me how to do "in-line" URL links. – Dave L. Renfro Dec 11 '20 at 18:43

1 Answers1

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This question gets into subtleties of how we define derivatives in terms of unidirecional limits. Let's consider the outcome in two different number systems.

Real analysis

The derivative exists iff two one-sided derivatives exist with equal value. The right-derivative$$\left.\lim_{h\to0^+}\frac{\sqrt{x+h}-\sqrt{x}}{h}\right|_{x=0}=\lim_{h\to0^+}\frac{\sqrt{h}}{h}=\color{blue}{+\infty}$$ is an extended real number; the left-derivative $\lim_{h\to0^-}\frac{\sqrt{h}}{h}$ does not exist.

Complex analysis

$f^\prime(0)$ exists with value $L$ iff $\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=L$ for all $\theta\in\Bbb R$. For $f(z)=z^{1/2}=e^{\tfrac12\ln z}$,$$\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=\lim_{r\to0^+}\frac{\sqrt{r}e^{i\theta/2}}{re^{i\theta}}=\color{red}{\infty}$$is an extended complex number. Do not confuse $\color{blue}{+\infty}$ with $\color{red}{\infty}$.

One more point: if we'd been differentiating $x^{1/3}$ at $x=0$, the two-sided derivative would again not exist in real analysis, but this time for a different reason: the left-derivative would exist, but would be $-\infty$ rather than $+\infty$.

J.G.
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  • OK I see but the right one sided derivative is equal to positive infinity so is it correct to say that the one sided derivative is defined as positive infinity – The homeschooler Dec 10 '20 at 10:03
  • @Thehomeschooler As long as "limit-is-infinity" statements are understood in their usual sense, yes. The definition of $\lim_{x\to a^+}g(x)=L$ isn't quite the same when $L=\infty$ as when $L\in\Bbb R$. – J.G. Dec 10 '20 at 10:10