Let $g=|x|$. Then, the derivative at $c=0$ is given by: $$ g'(0) = \lim_{x \to 0} \frac{|x|}{x} $$ which is either $+1$ if $x$ comes from the positive $x$-axis or $-1$ if $x$ comes from the negative $x$-axis. Now, it makes intuitive sense that the limit does not exists. But what is the rigorous reason?

I think it is due to the following theorem:

Let $g: A \to \mathbb{R}$, and let $c$ be a limit point of $A$. If there exists two sequences $(x_n)$ and $(y_n)$ in $A$ with $x_n \neq c$ and $y_n \neq c$, and: $$ \lim x_n = \lim y_n = c \;\;\; \text{and} \;\;\; \lim g(x_n) \neq \lim g(y_n) $$ then we can conclude that the functional limit $\lim_{x \to c} g(x)$ does not exist.

The logic is that since $\mathbb{Q}$ is dense in $\mathbb{R}$, we can always find a sequence $(x_n) \subseteq \mathbb{Q}$ that converges to zero from the positive $x$-axis with $x_n \neq 0$, and we can always find a sequence $(y_n) \subseteq \mathbb{Q}$ that converges to zero from the negative $x$-axis with $y_n \neq 0$. Thus, I believe we have: $$ \lim x_n = \lim y_n = 0 \;\;\; \text{and} \;\;\; \lim g'(x_n) = 1 \neq \lim g'(y_n) = -1 $$ Furthermore, $c=0$ is a limit point of $\mathbb{R}$. Thus, if this is true so far, then we can conclude from the above theorem that $\lim_{x \to 0} (|x|/x)$ does not exist.

Is the above reasoning/proof valid, or am I wrong?

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    $\begingroup$ It is correct, but $g'(0)$ is only equal to the given limit provided the limit exists; in this case, as you've shown, it does not. $\endgroup$ Commented Sep 5, 2014 at 13:55
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    $\begingroup$ this may help: math.stackexchange.com/questions/281601/… $\endgroup$
    – Alex
    Commented Sep 5, 2014 at 13:56
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    $\begingroup$ It is also obvious from the epsilon-delta definition. Your reasoning from the theorem is correct (tho at this point you don't need to invoke the density of the rationals; just exhibit two specific sequences, say $\pm1/n$; saying that you can "always" find such sequences, as you put it, is a little odd), but you might as well just start from the definition. $\endgroup$ Commented Sep 5, 2014 at 14:05
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    $\begingroup$ @Hunter: Travis is objecting to the first equation you wrote, setting $g'(0)$ equal to a limit. You can't write this equation until you know the limit exists. But in fact, the limit doesn't exist. So you can't write such an equation: it doesn't make any sense. (Well, it would if we allowed $g'$ to be a partial function, but we don't do that in analysis.) $\endgroup$ Commented Sep 5, 2014 at 14:08
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    $\begingroup$ @Hunter Yes, my remark is simply we shouldn't write $g'(0) = \cdots$ at all, given that $g'(0)$ is simply not defined, and hence not equal to anything. It's probably fastest just to compute the left- and right-handed limits and show they're different as thanasissdr helpfully has. If you're not convinced that this method is "rigorous", note that you can prove it is sufficient using essentially your argument in your original post. $\endgroup$ Commented Sep 5, 2014 at 14:09

1 Answer 1


Since there are multiple different approaches to defining limits, here is an explanation in terms of limits.

Theorem: Let $f$ be a function defined in a neighborhood of $c$ (though not necessarily at $c$). Then $\lim_{x\to c} f(x)$ exists if and only if both $\lim_{x\to c^+} f(x)$ and $\lim_{x\to c^-} f(x)$ exist and are equal.

Now, let $f(x)=\frac{|x|}{x}$. In this case, we have that the left sided limit exists and is $-1$ while the right sided limit exists and is $1$. Since these are unequal, the combined limit cannot exist.

Modulo a few details and some notation problems, I believe this is essentially the idea you were trying to express.


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