Proof: $f(x) = |x|$ is not differentiable at $x=0$. I am trying to prove that $f(x) = |x|$ is not differentiable at $x=0$. 
So far, 
I have that
$$\lim_{h \to 0} f(x) =$$
$$\lim_{h \to 0} (f(x+h)-f(x))/h =$$
$$\lim_{h \to 0} (|0+h|-|0|)/h =$$
$$|h-0|/h =$$
$$|h|/h$$ 
Now, I would like to prove that:
$$
g(x) = \left\{\begin{aligned}
&1 &&: x > 0\\
&-1 &&: x < 0
\end{aligned}
\right.$$
$$\lim_{x \to 0} g(x)$$ Does not exist
How would I do this using the definition of a limit, without using left/right handed limits.
Thanks!
 A: One way to do it is look at the sequence $g(x_n)$, where $x_n = (-1)^n/n$. This sequence is alternating and converges to $0$, and clearly an alternating sequence does not converge. Since the sequence converges, it has terms in any $\epsilon$ ball centered at $0$.
A: If we don't want to use derivative, we need to refer to the equivalent (more general) definition of differentiable function. 
Notably we say that a function is differentiable at $x=x_0$ if $\exists! L\in \mathbb{R}$ such that
$$f(x_0+h)=f(x_0)+L\cdot h+o(h)$$
In that case, since at $x=0$ $f(x)=|x|$ has a cuspid point, the tangent is not defined at that point and the function is not differentiable since we cannot find $L$ such that
$$f(h)=f(0)+L\cdot h+o(h)$$
indeed for $h>0$ we have
$$f(h)=1 \cdot h$$
but for $h<0$ we have
$$f(h)=-1 \cdot h$$

As an alternative, by the way you don’t like to use, we have that
$$\lim_{x\to 0^+}\frac{|x|-0}{x-0}=\lim_{x\to 0+}\frac{x}{x}=1$$
but
$$\lim_{x\to 0^-}\frac{|x|-0}{x-0}=\lim_{x\to 0+}\frac{-x}{x}=-1$$
therefore 
$$\lim_{x\to 0}\frac{|x|-0}{x-0}$$
doesn’t exist and $f(x)=|x|$ is not differentiable at $x=0$.
A: For $x\not=0$ we have $|g(x)-g(-x)|=2$.  Now if $\lim_{x\to0}g(x)=L$, then we would have $\lim_{x\to0}g(-x)=L$ as well, in which case we obtain the contradiction
$$2=|g(x)-g(-x)|=|(g(x)-L)+(L-g(-x))|\le|g(x)-L|+|g(-x)-L|\to0+0=0$$
A: We know that for any positive $h$, $\frac{|h|}{h}=1$ and for any negative $h$, $\frac{|h|}{h}=-1$.
Let $L$ be any candidate value for the limit. Let $d_1$ be the distance from $L$ to $1$ and $d_{-1}$ be the distance from $L$ to $-1$. 
There are two cases to consider:


*

*Suppose one of $d_1$ and $d_{-1}$ are $0$. Without loss of generality, we can assume $d_1=0$.
Then, for any interval around $0$, there will be negative values which take the value $-1$ which will be a distance of $1\gt\frac{1}{2}$.

*If neither $d_1$ or $d_{-1}$ is $0$ then let $\epsilon$ be half of the minimum of the two. Then, for any $x\ne 0$ and any $\delta\gt 0$ $|x-0|\lt\delta\implies |g(x)-L|$ is greater than or equal to the minimum of $d_1$ and $d_{-1}$ which is strictly greater than $\epsilon$. 

A: Can you not just show that
$$\lim_{x\to0}\Big[\frac{d}{dx}(-x)\Big]\neq\lim_{x\to0}\Big[\frac{d}{dx}(x)\Big]$$
therefore the derivative is undefined at that point?
