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!
 A: 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.
A: 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. 
