# Limits at Infinity and limit equality

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, this is quite a powerful trick that is often used to make certain limits more manageable to evaluate. Your reasoning for the validity of this technique is also pretty good! – Don Thousand Nov 29 '18 at 17:15
• I should note, however, that the reverse is not the case. If $\lim_{x\to\infty}f(x)=a$, then $\lim_{x\to0}f(\frac1x)$ is not necessarily $a$. However, $\lim_{x\to0^+}f(\frac1x)=a$ – Don Thousand Nov 29 '18 at 17:24

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

$$\lim_{x \to 0^+}f(g(x))=L$$