I'm given function $f:(a,\infty)\to\mathbb{R}$ which has a limit at infinity, i.e., $\lim_{x \to \infty}f(x)$ exists, call it $L$. And I want to show that given a function $g(x) := {f(1/x)},$ which is defined on $(0,1/a),$ that this function $g(x)$ has a limit at 0 if and only if the limit of $f$ as $x$ tends to infinity exists.
I know I have to use the $\epsilon - \delta$ defintion, but before that I think the following is an equivalent formulation:
\begin{gather}
\lim_{x \to \infty}f(x) = \lim_{x \to 0}f(1/x).
\end{gather}
I know this is just an exercise in chasing the $\epsilon - \delta$ notation, but I think the "trick" here is to use the fact that if $f$ has a limit at infinity, then for all $\epsilon > 0,$ there exists $M > a$ such that for all $x \geq M$ we have that $|f(x) - L| < \epsilon$. So I think the idea here is to pick my $\delta$ as $1/M$ since we have that
\begin{gather}
x \geq M \implies 1/x \leq 1/M
\end{gather}
and we know that if $x \geq M$ then $|f(x) - L| < \epsilon.$ So if we suppose $\epsilon_0 > 0$ and that $|f(1/x) - L| < \epsilon_0$ will $\delta_0 = 1/M$ suffice? My intuition says yes, but I am not sure how to formulate this rigorously.
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