# Showing the limit as $x \to 0$ of the function $x/|x|$ does not exist

How does one go about showing that $$\lim_{x \to 0} \frac{x}{|x|}$$ does not exist by using the definition of the limit, ie using epsilon and delta.

I imagine it's best to start off by attempting to find a contradiction by assuming that $$\lim_{x \to 0} \frac{x}{|x|}=L$$ for some L in the reals. Then there exists a ð>0 such that for all x in $$\mathbb{R}$$ excluding 0 with $$0<|x|<ð$$, $$|\frac{x}{|x|}-L|< something.$$ Obviously the function here is -1 for any x<0 and 1 for any x>0, but I'm not sure how to proceed from here.

How would I show a contradiction for this case?

Thanks!

Note that $$f(x)\equiv\frac{x}{\left|x\right|}=\begin{cases} +1 & \text{if }x>0,\\ -1 & \text{if }x<0. \end{cases}$$ The fastest way to establish the claim is to note that by the above, $$\lim_{x\uparrow0}f(x)=-1\neq+1=\lim_{x\downarrow0}f(x).$$ However, if you want to do it with a "direct" epsilon-delta argument, you can basically replicate the argument embedded in the above as follows:
Suppose $$L\equiv\lim_{x\rightarrow0}f(x)$$ exists. It follows that for $$\epsilon=1$$, we can find $$\delta>0$$ such that $$\left|f(x)-L\right|<1 \qquad \text{and} \qquad \left|f(-x)-L\right|<1$$ whenever $$x$$ is positive and strictly less than $$\delta$$. Simplifying these inequalities, we see that $$\left|L-(+1)\right|<1 \qquad \text{and} \qquad \left|L-(-1)\right|<1$$ (independent of $$\delta$$), a contradiction.
Let $$0<\epsilon <1$$Suppose $$|\frac x {|x|} -L| <\epsilon$$ for $$|x| <\delta$$. Take $$x=\delta /2$$ to get $$|1-L|<\epsilon$$. Then take $$x=-\delta /2$$ to get $$|-1-L|<\epsilon$$. So $$|1+L| <\epsilon$$. Now $$2=|2|=|(1+L)+(1-L)|\leq |1+L|+|1-L| <\epsilon+\epsilon <1+1=2$$ which is a contradiction.