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A disagreement in my calculus class has arisen as to whether $f(x) = \frac{x}{x}$ is differentiable for the domain of all real numbers, including $0$.

According to our textbook, for a function to be differentiable at $a$, it must be continuous at $a$. What we're not sure about is:

  1. Is $f(x)$ actually discontinuous at $0$?
  2. If it's discontinuous at $0$, how come we can still obtain a derivative of $0$ by hand using limits? Mathematica seems to agree with $f'(x) = 0$.

Edit: To clarify, the specific function in question is $f(x) = \frac{x}{x}$

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    $\begingroup$ No, it's not discontinuous at $x=0$ because it's not defined there. For the same reason, it's not differentiable there. $\endgroup$ – MathematicsStudent1122 Oct 17 '16 at 13:38
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    $\begingroup$ Your problem has arisen because of a serious failing in the teaching of this material at the basic level. The texts and most of the teachers refuse to give due importance to the concept of domain of a function, and allow a function to be “defined” by a formula, without mention of the domain. In fact, the definition of any function is incomplete without specification of the domain. $\endgroup$ – Lubin Oct 17 '16 at 18:12
  • $\begingroup$ @Lubin Yes, the definition of domain I've heard--both in college and high school--is very different from what's made apparent here. I was previously taught that domain was something arbitrarily defined by the mathematician, rather than a property of the function--and all of my textbooks and professors continue to stand by this definition. This has never made much sense to me and is probably the cause of this misunderstanding. $\endgroup$ – Zenexer Oct 22 '16 at 3:37
  • $\begingroup$ Yes and no. A naked formula will have a “natural” domain associated to it. In this case, if you accept the natural domain, you must still remember the significance of the domain: continuity is only even defined at points of the domain. $\endgroup$ – Lubin Oct 22 '16 at 17:47
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This highly depends on how you are using the formula $\dfrac{x}{x}$ to define your function $f$.

To be precise $f(x)=\dfrac{x}{x}=1$ when $x \not=0$ and is undefined when $x=0$. Because $f$ is not defined at $x=0$ it is not continuous there. Also, because it is not defined at $x=0$, its derivative is not defined at $x=0$. Consider the difference quotient: $\dfrac{f(0+h)-f(0)}{h}$. In this difference quotient "$f(0)$" is undefined (so you can't even begin to compute the limit).

That said, $x=0$ is a removable singularity. So we can continuously fill in the value $f(x)=1$ when $x=0$ and "repair" our function. Using the new formula: $f(x)=1$ (which now gives a function defined on the whole real line) we get a function differentiable everywhere. When Mathematica tells you that $f$ is differentiable at $x=0$, it is removing the singularity for you.

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    $\begingroup$ Would it be correct to say that 0 isn't in the domain of the function? $\endgroup$ – Zenexer Oct 17 '16 at 13:46
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    $\begingroup$ If you use the formula $x/x$, then yes, $x=0$ is not in the domain. However, you have to be careful. Many people automatically remove that kind of singularity (they decide for you that you've used a "bad" formula and "fix" it for you) in which case they'd say that $x=0$ is in the domain. $\endgroup$ – Bill Cook Oct 17 '16 at 13:56
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The real question, you need to answer first is:

How do you define $f$ at $0$?

Up to now, I don't see some $f$ defined at $0$. For a function to be continuous (let alone differentiable) at some point you need it to be defined at that point.

So:

  • If you define $f(0)=1$, the function is constant, hence differentiable, hence continuous.
  • If you define $f(0)$ different, the function will not be continuous, hence not differentiable.

What one may say in this situation is:

The function is not defined at $x=0$ but for neighborhoods $U$ of $x$, it is defined on $U\setminus\{x\}$, and the limit $\lim_{x\to 0}f(x)$ is well defined (i.e. independent of how $x$ approaches zero), hence

$f$ can be extended continuously to $x=0$.

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  • $\begingroup$ $f(x) = \frac{x}{x}$ $\endgroup$ – Zenexer Oct 17 '16 at 13:39
  • $\begingroup$ Err, no. That does not make sense for $x=0$… $\endgroup$ – Dirk Oct 17 '16 at 13:40
  • $\begingroup$ Right. Hence the question, and why we're confused. $\endgroup$ – Zenexer Oct 17 '16 at 13:42
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    $\begingroup$ Be confused no more! $\endgroup$ – Dirk Oct 17 '16 at 13:42
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The problem is your definition of $f$. If $f$ is the function defined by $f(x)=1$, then of course the derivative exists everywhere and is equal to zero.

If, however, you define the function $f$ as taking an input $x$ and dividing it by $x$ (i.e., $f(x)=x/x$), this function is technically undefined at zero (what does it mean to divide by $x=0$?). Therefore, you cannot talk about its derivative at zero, since one normally defines derivatives on open sets where the function is itself defined. However, you can show that $f^\prime(x)=0$ everywhere other than zero, and hence $f^\prime(x)\rightarrow 0$ as $x\rightarrow 0$.

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The function isn't continuous at $0$. It's not discontinuous either. It just isn't, and that's the end of it. That being said, there exactly one nice, everywhere continuous and differentiable function defined on all of $\Bbb R$ that coincides with your $f$ on the domain of $f$.

Extending the domain of $f$ in this fashion is called removing a singularity, and isolated points where a given function is not defined, but where you can extend nicely in this fashion is called a removable singularity.

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$$f(x)=\frac{x}{x}=\begin{cases}1 &, x \not=0 \\ \text{undefined} &, x=0\end{cases}$$ Yes, the function $f(x)=\frac{x}{x}$ is not defined at $x=0$. The right and left-hand limits of the function at $x=0^-$ and $x=0^+$ exist finitely and are equal to $1$. But $f(0)$ is undefined and hence you cannot make a proper mathematical statement involving $f(0)$, $\lim_\limits{x\to0^-}f(x)$ and $\lim_\limits{x\to0^+}f(x)$.

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    $\begingroup$ No, the function isn't discontinuous at $0$. It is simply not defined, and that's that. $\endgroup$ – Arthur Oct 17 '16 at 13:40
  • $\begingroup$ @Arthur If you were to draw the graph of this function, could you do so without lifting your pen/pencil? The domain being $\mathbb{R}$? A very basic trick to check continuity? $\endgroup$ – SchrodingersCat Oct 17 '16 at 13:44
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    $\begingroup$ But that's not the definition. Don't confuse basic tricks and heuristics for facts. If you apply the $\epsilon$-$\delta$ definition, or the inverse-image-of-open-sets definition, then asking whether $f$ is continuous or not at $0$ simply doesn't make sense. $\endgroup$ – Arthur Oct 17 '16 at 13:45
  • $\begingroup$ Can you draw the function $f:\mathbb{N}\to\mathbb{N}$, $f(n)=n$ without lifting you pencil? Can you draw the Weierstrass or the Takagi function (with or without lifting your pencil)? $\endgroup$ – Dirk Oct 17 '16 at 13:47
  • $\begingroup$ @Dirk They are pathological functions.. they cannot be drawn by hand ... and don't go for complicated meanings please.. What I meant to ask has been well understood and answered by Arthur. And BTW $f(n)=n$ can be drawn without lifting the pencil, provided you give me sufficient paper and ink (and money for spending my precious time). $\endgroup$ – SchrodingersCat Oct 17 '16 at 13:51

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