Equivalence of limit at infinity and one-sided limit at zero Let $f$ be a real-valued function defined in $(0,a)$ for some $a>0$. Are the following two equivalent?
$$\lim_{x \to 0 \atop x>0} f(x) = l$$
$$\lim_{x \to +∞} f\left(\frac{1}{x}\right) = l$$
This (and the equivalent for $x<0$ and $-∞$) seems to be assumed in the proofs of some theorems (for example the derivative of $\ln(x)$). I looked for a theorem stating this formally in wikipedia and in several books, but I didn't find such proposition.
 A: The theorem might say the following:

Suppose $g$ is continuous at $a$ and $g(x)$ differs from $g(a)$ for $x$ sufficiently close to $a$.  Further suppose $\lim\limits_{u\to g(a)} f(u)$ exists.  Then
  $$
\lim_{x\to a} f(g(x)) = \lim_{u\to g(a)} f(u)
$$

In this case you would have $g(x) = \dfrac 1 x$ and $a=+\infty$.
You could regard $+\infty$ as a point in a space $\mathbb R\cup\{+\infty,-\infty\}$ and regard intervals $(m,+\infty]$ as its open neighborhoods and write a proof in the language of topology.  Or you could write an $\varepsilon$-$\delta$ proof.
A: Yes they are equivalent. Let's consider one direction: Suppose $\lim_{x\to 0^+}f(x) = L.$ We want to show $\lim_{x\to \infty}f(1/x) = L.$
Let $\epsilon>0.$ Then there exists $\delta >0$ such that $0 < x < \delta$ implies $|f(x)- L| <\epsilon.$ Set $M = 1/\delta.$ Then $x> M$ implies $0 < 1/x < 1/M = \delta.$ Hence $x> M$ implies $|f(1/x)- L|< \epsilon.$ This is what we needed to show.
I'll let you handle the converse; it's very much the same and will be more valuable to you if you do it than if I write and you nod.
