Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a monotone function such that $\displaystyle{\lim_{n\to\infty}}f(n) = L$. Then, $\displaystyle{\lim_{x\to\infty}}f(x) = L$.
I know that if $\displaystyle{\lim_{x\to\infty}}f(x) = L$, then $\displaystyle{\lim_{n\to\infty}}f(n) = L$.
Let $\varepsilon>0$. Since $\displaystyle\lim_{n\to\infty}f(n)=L$, there exists $M$ such that if $n>M$ and $n$ is an integer, then $|f(n)-L|<\varepsilon$.
Now let $x$ be (a real number) greater than $M$. Suppose $f$ is monotonic increasing (the decreasing case is similar). Then $f(x)\ge f(M)$. Suppose $f(x)>L$. Then as $f$ is increasing, $$f(\lceil x\rceil)>L$$ so $$\lim_{n\to\infty}f(n)>L\ ,$$ that is, $L>L$, a contradiction. Hence $f(x)\le L$. We have shown that $$f(M)\le f(x)\le L$$ and so $$|f(x)-L|=L-f(x)\le L-f(M)=|f(M)-L|<\varepsilon\ .$$ We have shown that for all $\varepsilon>0$ there exists $M$ such that if $x>M$ then $|f(x)-L|<\varepsilon$. By definition, $\displaystyle\lim_{x\to\infty}f(x)=L$.