Prove that if $\lim_{x\to\infty}f(x)=L$ exists and finite, and $\lim_{x\to\infty}\lfloor f(x)\rfloor$ doesn't exist then L is an integer 

Let $f$ be a continuous function on $(0, \infty)$ s.t $\lim \limits_{x \to \infty}f(x) = L$ exists and finite, but $\lim \limits_{x\to \infty} \lfloor f(x) \rfloor$ doesn't exist. Prove that L is an integer .


I tried to use the fact that for each $\epsilon >0$ there is $x_o$ that for each $x>x_0$ : $|f(x)-L|< \epsilon$. and suppose negatively that L is not an integer and get a contradiction.
Thats by choosing such an epsilon that guarantees that $\lfloor f(x)\rfloor=\lfloor L\rfloor$ when $x>x_0$, or any other combination that tells us $\lim \limits_{x \to \infty}\lfloor f(x)\rfloor = \text{exists}$
I didn't know how to continue from here, or find such an epsilon.
NOTE: $\lfloor f(x)\rfloor$ is the floor function. 
 A: I am going to do a proof by contradiction. Assume the opposite, that $L$ is not an integer. Note that because $L$ is not an integer, $\lfloor L\rfloor<L<\lceil L\rceil$. Because $f(x)\to L$ as $x\to\infty$, we can choose $\varepsilon=\min\{L-\lfloor L\rfloor, \lceil L\rceil-L\}$ such that there exists an $M$ such that for all $x>M$, $|f(x)-L|<\varepsilon$.
If $L-\lfloor L\rfloor\le \lceil L\rceil-L$, then $\varepsilon=L-\lfloor L\rfloor$ and
$$|f(x)-L|<\varepsilon\\\Rightarrow\lfloor L\rfloor-L<f(x)-L<L-\lfloor L\rfloor\le\lceil L\rceil-L\\
\Rightarrow\lfloor L\rfloor<f(x)<\lceil L\rceil$$
If $L-\lfloor L\rfloor> \lceil L\rceil-L$, then $\varepsilon=\lceil L\rceil-L$ and
$$|f(x)-L|<\varepsilon\\\Rightarrow \lfloor L\rfloor-L<L-\lceil L\rceil<f(x)-L<\lceil L\rceil-L\\
\Rightarrow\lfloor L\rfloor<f(x)<\lceil L\rceil$$
And so, for all $x>M$, $f(x)\in(\lfloor L\rfloor,\lceil L\rceil)$. Then $\lfloor f(x)\rfloor = \lfloor L\rfloor$ for all $x>M$, and $\lim_x\lfloor f(x)\rfloor$ exists and is equal to $\lfloor L\rfloor$.
