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How can I proof that the following function is not differentiable at $x = 3$ using the definition of a derivative?

$f: (0,2) \cup \{3\} \to \mathbb{R}$ with $f(x) = 5$

My attempt was as follows:

$$f'(3) = \lim_{w \to z} \frac{f(w) - f(z)}{w - z} = \lim_{w \to z} \frac{5 - 5}{w - 3} = 0$$

It seems to me as if the definition I used doesn't "recognize" that I cannot approach $3$ from both sides, given that $3$ is an isolated point. Should I have used another definition instead? Or am I missing a fundemantel step?

Soham Saha
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Music and Chill
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jan 18 '25 at 13:05
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    You should include the definition of "differentiable" you want applied, as nearly all definitions (indeed, perhaps all definitions, if we're talking about "ordinary" notions of differentiability) I'm familiar with require the point being investigated to be a limit point of function's domain (in addition to belonging to the function's domain). – Dave L. Renfro Jan 18 '25 at 13:09
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    @JoséCarlosSantos Thank you for the helpful comment. I am going to edit my question and provide more context and my attempt of proving it as soon as I figure out how to use MathJax properly. – Music and Chill Jan 18 '25 at 13:17
  • It will help to explain why you are interested in finding the derivative of this function at the given point, or for that matter, the derivative of any function at a point in the function's domain that is an isolated point relative to the function's domain. For example, did someone (friend you know in person, internet stranger, your teacher, etc.) ask you, did you see this problem in a textbook (if so, specify the book, edition, year, author, page number), on a web site somewhere (if so, provide the web site URL), or somewhere else (e.g. something you just made up and were curious about)? – Dave L. Renfro Jan 18 '25 at 13:47
  • @Dave L.Renfro: At my university I notice that my professor often assumes that f is a function with an intervall I ⊆ R with no isolated points and I wanted to understand, why exactly we are assuming that the intervall I does not have any isolated points. So I just thought of an example to proof that a function is not differentiable at an isolated point (of course not generally, only bound to the example I have shown in the question). – Music and Chill Jan 18 '25 at 14:01
  • @DaveL.Renfro: or let me ask this question another way: How do we know that a function f is not differentiable at isolated points? Intuitively it makes absolutely sense, because we cannot approach that point from either side, which is required for differential calculus to operate. However, I don't know how to proof that, even with the little example I have thought of in the question above. – Music and Chill Jan 18 '25 at 14:04
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    How do we know that a function f is not differentiable at isolated points? As far as I'm aware, there is no use for defining what it means for a function to be differentiable at an isolated point, so pretty much all definitions I've seen explicitly state that the point is not isolated, at least when the discussion is sufficiently nuanced that the issue needs to be considered. Indeed, typically the function is assumed to be defined in a neighborhood of the point, although this is slightly lessened in certain contexts (e.g. one-sided neighborhoods for left/right derivatives). – Dave L. Renfro Jan 18 '25 at 14:44

1 Answers1

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As an answer to your concern raised in the comments:

Given a subset $E \subseteq \mathbb{R}$ and a point $x \in E$, if $x$ is not an isolated point of $E$ we call it a limit point of $E$, and we require in the definition of the derivative that $x$ be a limit point of $E$ for $f'(x)$ to exist. We choose this definition as, intuitively, differentiability is concerned with the local behaviour of a function, and for isolated points, there is an interval $(x_0-\epsilon, x_0+\epsilon)$ for some $\epsilon > 0$ such that the function is only defined at $x_0$ - hence we cannot talk about local behaviour.

More concretely, let $E = \{0\} \cup \{\frac 1n : n \in \mathbb{N}\}$, and define $f : E \to \mathbb{R}$ by $f(x) = x$. Then, it is a good exercise in definitions to show that 0 is a limit point of $E$, and hence $f'(0) = 1$, whereas every other point of $E$ is isolated.

altwoa
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