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Does existence of a limit at a point not necessarily mean that it's differentiable at that point?

Take this function:

$$f(x) = \frac{(x - 1)^{2}}{x - 1}$$

The function is not defined at x = 1, but as x approaches 1, f(x) goes to 0; i.e., a "removable discontinuity." But the function is not differentiable at 1, right?

calc_student
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  • If you define $g(1)=0$, $g(x) = \frac{(x-1)^2}{x-1}$ for $x \neq 1$ then $g(x)=x-1$ for all $x$ and this is the function with the discontinuity removed. Moreover, $g'(x)=1$. So, once the discontinuity is removed it's everywhere differentiable. But, I think differentiability has little to do with the heart of your question. – James S. Cook Aug 10 '13 at 14:46

5 Answers5

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You're right. $f(x)$ as written has a discontinuity at $x=1$, so it is not considered to be differentiable. However, if you modify $f(x)$ by inserting the removable discontinuity, $f(x)$ becomes a differentiable function. That is $$ g(x)=x-1 = \begin{cases} \frac{(x-1)^2}{x-1} & x\neq 1\\ 0 & x = 1 \end{cases} $$ is a differentiable function

As you said, the existence of a limit at a point (even along with differentiability in the neighborhood of that point) does not guarantee that the function is differentiable at that point. In order for a function to be differentiable at a point, it must first be continuous at that point.

Related: differentiability implies continuity.

Community
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Ben Grossmann
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  • No, the function is continuous everywhere (on its domain). There is no point of talking about whether a function is continuous/differentiable at x when that x is not in the domain of the function. – scibuff Dec 04 '16 at 16:37
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    @scibuff it depends on the context. Any calculus teacher/textbook (in the USA) will tell you that gaps in the domain of a function count as discontinuities, and that given a function $f(x)$, one may ask "what is the domain of the function $f$?" Of course, neither of these things are true in the language of modern mathematics; domains must be specified a priori and a function is necessarily defined at every point in its domain. It comes down to different usage. – Ben Grossmann Dec 04 '16 at 17:35
  • I completely disagree with that. The very definition of continuity (epsilon-delta) requires the function to be defined at the point where it is continuous/discontinuous.

    The misconception comes from the notion that continuous functions are those which you can draw (graph) without having to lift the pen off the paper. That is, indeed, incorrect as there are continuous functions one cannot even graph, e.g. Thomae's function.

    p.s. I'd think Spivak's calculus is easily accessible in the US

     – scibuff Dec 04 '16 at 20:48
  • I am very well aware of the definitions you are using. These definitions are quite distinct from those likely used in OP's class. – Ben Grossmann Dec 04 '16 at 21:26
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Continuity is necessary for the existence of a $f'$, that means if a function $f$ has discontinuity at $x=a$ then $f$ is not differentiable at $x=a$. Your function is not continuous at $x=1$, therefore it's not differentiable at $x=1$. Moreover, continuity is not a sufficient condition.For example $|x|$ is continuous over $\mathbb{R}$ but it's not differentiable at $x=0$.

user66733
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For the example that you have given, the function is not differentiable at that point because of the discontinuity that exists at that point.

Remember that the definition of a derivative at point uses a limit e.g.

$$f' (a) = \lim_{h\rightarrow 0}\frac{f(a+h) - f(a)}{h}$$

Thus, it is only differentiable when a limit exists so yes, you are right when you say that if a limit exists, the function is differentiable and when it doesn't exist, the function is not differentiable.

Jeel Shah
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  • @downvoter If you would could help me improve this post, that would be great and explain your downvote too. – Jeel Shah Aug 10 '13 at 15:01
  • Your answer is wrong, and thus detracts from the question. $f(x)$, as defined in the question, is not considered differentiable. As confusing as that may be, that is how the definition works. – Ben Grossmann Aug 10 '13 at 15:04
  • If you remove the discontinuity, the function is differentiable. It's simple, graph the function and you'll see for yourself that it's differentiable. If you take a function that is not continuous then the function won't be differentiable but this function is continuous. – Jeel Shah Aug 10 '13 at 15:06
  • I understand that the function is continuous if you fill in the discontinuity. That much is clear. However, as long as the function is missing that point, it is discontinuous, which is why we call it a "discontinuity". Since the function is not continuous at that point, it is not differentiable. – Ben Grossmann Aug 10 '13 at 15:09
  • Okay, that makes sense. I will change the post to reflect that. Thanks! – Jeel Shah Aug 10 '13 at 15:13
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By the very definitions of these concepts, a function is continuous only at points $a$ where it is defined ($\lim_{x\to a} f(x)=\color{red}{f(a)}$) and it is differentiable only at points $a$ where it is defined ($\lim_{x\to a}\frac{f(x)-\color{red}{f(a)}}{x-a}$ exists). We also find as a simple theorem that functions are continuous where they are differentiable, but for the problem at hand it suffices to note that $f$ is not even defined at $x=1$, hence not differentiable there.

One must distinguish between the given function $f$ and the function obtained by removing the removable singukarity (which is of course continuous and in this case also differentiable at th epoint in question)

Hagen von Eitzen
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I think it depends on the domain of the function.

  1. if the domain is $\mathbb{R}$, then $f(x)$ is not defined at $x=1$, so it's not continuous and not differentiable;

  2. if the domain is $\mathbb{R}$ but $x\ne 1$, then it's continuous and differentiable in its domain.

Siong Thye Goh
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carolynzy
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