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?
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.