How to show that $f(x)=|x|/x$ does not have any limit as $x\to0$?

You just have to find one point $x_0$ in $(-\delta, \delta)$ such that $|f(x_0) - L| \geq 1/2$. Let's say we first suppose that $L \geq 0$ -- could you then find a point that would do the trick?

If $L \ge 0$ we can choose $x\in(-\delta/2,0)$ so that $|f(x)-L| = |-1-L| = L+1 > 1/2$
If $L < 0$ we can choose $x \in (0,\delta/2)$ so that $|f(x)-L| = |1-L| > |1-0| = 1 > 1/2$

You have done all the thing. Now if $L$ is greater than zero choose a $x<0$. Then you shall have $|f(x)-L|= |1+L|>\frac{1}{2}$ and similarly for $L<0$ choose a $x>0$ and you will have the result using similar arguments.