Can we replace the limit of a sequence with that of a function? Let $f$ be a function defined in $[1,\infty]$. If $\lim_{x\to\infty}f(x) =
L$ and $a_n = f(n)$ for integer $n\ge 1$ then $\lim_{n\to\infty}a_n = L$.
Found this theorem in many references, but couldn't prove it or find the proof.
 A: Let $\varepsilon>0$. Then $\exists M\geq 1$ such that $\forall x\geq M$, $\vert f(x)-L\vert < \varepsilon$. This follows from the fact that the limit of $f(x)$ as $x\rightarrow +\infty$ is $L$.
Now let $N=\left \lceil{M}\right \rceil $ (integer part of $M$ + 1).Then $\forall n\geq N$, $\vert f(n)-L\vert\ < \varepsilon$.
So $\forall \varepsilon > 0$, $\exists N\in\mathbb{N}$, such that $\forall n\geq N$, $\vert a_n-L\vert < \varepsilon$.
A: The definition of "$\lim\limits_{x\longrightarrow\infty}f(x)=L$" that you're probably having in mind is
$\forall\epsilon >0\;\exists x(\epsilon)>0\;\forall x\geq x(\epsilon):\;\left| f(x)-L\right|<\epsilon\;\;\; (\ast)$
This definition admits a sequential characterization: You can show that it is equivalent to
$\forall (x_n)\subset\mathbb{R}:\; x_n\longrightarrow\infty\;\Rightarrow\; f(x_n)\longrightarrow L$.
whence your question is just the special case $x_n=n$.
To prove the equivalence (or just the direction you need) just write out the standard definitions of divergence $x_n\longrightarrow\infty$ and convergence $a_n=f(x_n)\longrightarrow L$. It should then become obvious how the sequential version follows from $(\ast)$.
