Let $f: [1,\infty) \rightarrow R$ be a function. Suppose that f is increasing, prove that if $\lim_{x \rightarrow \infty} f(x)$ exists then the sequence $\{f(n)\}_{n=1}^{\infty}$ is convergent.
So by definition of limit of function at infinity, let $\lim_{x \rightarrow \infty} f(x)=L$, then for any $\epsilon>0$, there exists $M \in R$ such that for $x \in [1,\infty)$, $x>M$, we have $|f(x)-L|<\epsilon$. I want to connect this to definition of limit of sequence: for any $\epsilon>0$, there exists $N \in \mathbf{N}$ such that $n>N$ implies $|f(n)-L|<\epsilon$.
It seems pretty easy, can I just let $N=M$, and $x=n$ (I have one concern though: $M \in R$, but N needs to be in $\mathbf{N}$.
Thanks!