I couldn't find the answer neither on Google, nor this website, so decided to ask.
The Heine definition of limit: from Wikipedia
$\lim_{x\to a}f(x)=L$ if and only if for all sequences $x_n$ (with $x_n$ not equal to $a$ for all $n$) converging to $a$ the sequence $f(x_n)$ converges to $L$.
Does it work with limit of a function at infinity, using a sequence that converges to infinity? (for example: $x_n = n$)
Thank you.
Best Answer
Yes.
Suppose $\lim_{x\to+\infty}f(x)=L$. That is, for every $\epsilon>0$ there exists $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Let $x_n\to+\infty$, that is for every $M>0$ exists $n_0\in\Bbb N$ such that $n>n_0$ implies $x_n>M$. I claim that $f(x_n)\to L$. Indeed, let $\epsilon>0$ be given. Then pick $x_0$ such that $x>x_0$ implies $|f(x)-L|<\epsilon$. Then pick $n_0\in\Bbb N$ such that $n>n_0$ implies $x>x_0$. It follows that $|f(x_n)-L|<\epsilon$ for $n>n_0$.
Next suppose $\lim_{n\to\infty}f(x_n)=L$ for all sequences with $x_n\to\infty$. I claim that $\lim_{x\to+\infty}f(x)=L$. Indeed, let $\epsilon>0$ be given. Assume there does not exist $M$ such that $x>M$ implies $|f(x)-L|<\epsilon$. Then we can define a sequence $x_n$ as follows: For given $n\in \Bbb N$ pick $x_n$ arbitrary with $x_n>n$ and $|f(x_n)-L|\ge\epsilon$ (such $x_n$ exists by our assumption applied to $M=n$). Then clearly $x_n\to +\infty$ and hence $|f(x_n)-L|<\epsilon$ for sufficiently large $n$. As this contradicts $|f(x_n)-L|\ge\epsilon$, the initial assumption must be wrong. That is: There does exist some $M$ such that for all $x>M$ we have $|f(x)-L|<\epsilon$.
Alternatively, observe that $x\to+\infty$ is equivalent to $\frac1x\to0^+$ and use that to translate the desired statement about $+\infty$ to a statement about finite $a$.