Prove that $\lim_{x \to 0} \frac{x}{|x|}$ doesn't exist using epsilon delta definition I'm trying to prove that $\displaystyle\lim_{x \to 0} \dfrac{x}{|x|}$ doesn't exist using epsilon-delta definition. I know i'm supposed to assign a value to epsilon and assume $\displaystyle\lim_{x \to 0} \dfrac{x}{|x|} = L$ and work backwards to get delta. But I'm now stuck on finding delta.
 A: First note that the function you are considering is:
$$
f(x) = \frac{x}{\lvert x \rvert} = \begin{cases}1 & \text{for }x>0 \\ -1 & \text{for } x<0\end{cases}.
$$
(So the function is not defined at $0$.)
Now let $\epsilon= \frac{1}{2}$ be given. You want to prove that there does not exist an $L\in\mathbb{R}$ and a $\delta > 0$ such that if $\lvert x \rvert < \delta$ then $\lvert f(x) - L\rvert < \epsilon = \frac{1}{2}$. Assume that such did exist. So then for $x\in (-\delta , \delta)$, you would have
$$
\lvert f(x) - L\rvert < \epsilon = \frac{1}{2}.
$$
That is, you have for all $x\in (-\delta, \delta)$ that 
$$
-\frac{1}{2} < f(x) - L < \frac{1}{2}.
$$
In particular for $x= \pm\frac{\delta}{2}$ you get the two equations
$$
-\frac{1}{2} < -1 - L < \frac{1}{2} \\
-\frac{1}{2} < 1 - L < \frac{1}{2}.
$$
So you get
$$
\frac{1}{2} < - L < \frac{3}{2} \\
-\frac{3}{2} <  -L < -\frac{1}{2}.
$$
But that obviously doesn't work.
Here is my challenge to you: redo this with $\epsilon = 1$.
A: Hint: $|x|=-x$ when $x<0$ and $|x|=x$ when $x>0$. Now calculate the limit from the left and from the right. Do they match? What does this imply?
