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.
Yes that's completely fine we have indeed
$$\lim_{y \to \infty}f(y) =L \iff \forall \epsilon>0 \quad \exists \bar y>0 \quad \forall y> \bar y \quad |f(y)-L|<\epsilon$$
and since for $g(x)= \frac1x$ we have
$$\lim_{x \to 0^+}g(x) =\infty \iff \forall M>0 \quad \exists \delta>0 \quad \forall x>0 \quad x<\delta \quad g(x)> M$$
then by $\bar y =M$ for $f(g(x))$ we have that
$$\forall \epsilon>0 \quad \exists \delta>0 \quad \forall x>0 \quad x<\delta \quad |f(g(x))-L|<\epsilon$$
that is
$$\lim_{x \to 0^+}f(g(x))=L$$