How to show that $f(x)=|x|/x$ does not have any limit as $x\to0$? 
$f(x)$ does not converge to any $L$ as $x\to a$ if for every $L$ there is $\epsilon>0$ such that for all $\delta>0$ there is $x$ such that $0<|x-a|<\delta$ and $|f(x)-L|\geq\epsilon$.

I wish to prove that
$$
f(x)=\frac{|x|}{x}=\begin{cases}
1,&x>0\\
-1,&x<0
\end{cases}
$$
does not converge to any $L$ as $x\to0$ using the above definition.
This is what I did: Fix $L$ and take $\epsilon=\frac{1}{2}$. For any $\delta>0$ there is $x$ such that $0<|x-0|<\delta$ and
$$
|f(x)-L|=\begin{cases}
|1-L|,&x>0\\
|-1-L|,&x<0
\end{cases}=
\begin{cases}
|1-L|,&x>0\\
|1+L|,&x<0
\end{cases}
$$
but I am not sure how I should show that $|f(x)-L|\geq\frac{1}{2}=\epsilon$.
 A: 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?
A: 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 $|f(x)-L| = |1-L| > |1-0| = 1 > 1/2$
A: 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.
A: Using Sequences:
$\lim_{x \rightarrow a} f(x) = c$ if 
any sequence $(x_n)_{n\in \mathbb{N} }$ with 
$ \lim_{n \rightarrow \infty } x_n = a$ implies
$\lim_{n\rightarrow \infty} f(x_n) = c$.
Choose:
$x_n = 1/n$ and  $y_n = - 1/n$, where 
$n \in \mathbb{N^+}$.
$\lim_{n \rightarrow \infty} x_n =$ $ \lim_{n \rightarrow \infty} y_n  = 0$;
$\lim x_{ n  \rightarrow \infty} f(x_n) = 1$;
$\lim y_{ n \rightarrow \infty} f(y_n) = - 1$.
Limit does not exist.
A: By contradiction: Suppose $L=\lim_{x\to 0}f(x).$ Then $$\forall s>0\;\exists r>0 \;\forall x\; (\;|x|<r\implies |f(x)-L|<s\;)\implies$$    $$\implies \exists r>0\;\forall x\;(\;|x|<r\implies |f(x)-L|<1\;)\implies$$    $$\implies \exists r>0\;(\;|f(r/2)-L|<1\;\land \;|f(-r/2)-L|<1\;) \implies$$   $$\implies \exists r>0\; (\;2=|f(r/2)-f(-r/2)|=$$ $$=|(f(r/2)-L)+(L-f(-r/2)|\leq$$ $$\leq |f(r/2)-L|+|L-f(-r/2)|<1+1\;) \implies$$ $$\implies \exists r>0\;(2<2).$$ 
