disproving smallest integer function limit prove that the limit of this function does not exist : 
$\lim_{x \to 1} \lfloor x \rfloor$ 
I know that the value of $\lfloor x \rfloor$ when $ x \to 1^-$  is $0$,
the value if  $\lfloor x \rfloor$ when $ x \to 1^+$  is $1$.
What is the approach of proving that a limit does not exist using the definition of the limit?
I started with assuming that the limit for  $\lfloor x \rfloor$ exist,
$$
|\lfloor x \rfloor - L| < \epsilon 
$$
$$
L - \epsilon <\lfloor x \rfloor < L+ \epsilon
$$
I chose $\epsilon = 0.5$
$$
L - 0.5<\lfloor x \rfloor < L+ 0.5
$$
but didn't know how to make a contradiction to prove it
Some help?
Thanks!!
 A: Let our function be $f(x)$. Suppose that the limit as $x\to 1$ of $f(x)$  exists, and is equal to $b$. Then for every $\epsilon\gt 0$, there is a $\delta$ such that if $0\lt |x-1|\lt \delta$, then $|f(x)-b|\lt \epsilon$. We will use this to obtain a contradiction.
Let $\epsilon=1/10$, and suppose that $|f(x)-b|\lt \epsilon$ whenever $0\lt |x-1|\lt \delta$. 
There are $x$ such that $0\lt |x-1|\lt \delta$ and $f(x)=1$. (Just let $x=1+\delta/2$.) So $|1-b|\lt 1/10$
Similarly, there are $x$ such that $0\lt |x-1|\lt \delta$ and $f(x)=0$. (Just pick $x=1-\delta/2$.) So $|0-b|\lt 1/10$.
But we have reached a contradiction, since there is no $b$ which is simultaneously within $1/10$ of $1$ and within $1/10$ of $0$. 
If you prefer, you can instead show that $b$ can be neither $\le 0.5$ nor $\gt 0.5$. It is for example not $\le 0.5$ because if it were we would have $|1-b|\ge 0.5$, which contradcts the fact that $||-b|\lt 1/10$.
A: If by the "definition of the limit" you mean the epsilon-delta definition, it's pretty straightforward.
Choose $\varepsilon$ so $0<\varepsilon<1$ Then, for any $\delta>0$, $\lfloor1-\frac{\delta}{4}\rfloor=0$ and $\lfloor1+\frac{\delta}{4}\rfloor=1$, and $|1-0|>\varepsilon$.
Therefore, the limit does not exist.
