Problem 81(a) from Section 2.6 of James Stewart's Calculus: Early Transcendentals (8e) asks us to prove that
$$\lim\limits_{x\to\infty}f(x)=\lim\limits_{x\to 0^{+}}f\left(\frac{1}{x}\right)$$
provided that these limits exist. After thinking for a while, I convinced myself that this result can be strengthened to
$$\lim\limits_{x\to\infty}f(x)=L\iff \lim\limits_{x\to 0^{+}}f\left(\frac{1}{x}\right)=L$$
This does indeed strengthen the result because (1) it only requires the existence of at least one of the limits in question, and (2) the existence of either limit implies the existence of the other, and thus implies that $\lim_{x\to\infty}f(x)=\lim_{x\to 0^{+}}f(1/x)$. Here's my attempt:
Let's first prove that $\lim_{x\to\infty}f(x)=L\implies \lim_{x\to 0^{+}}f(1/x)=L$. Since $\lim_{x\to\infty}f(x)=L$, we have that for all $\varepsilon >0$, there exists a $\delta$ such that $x>\delta\implies |f(x)-L|<\varepsilon$. Since $\lim_{x\to 0^{+}}1/x=\infty$, there exists a $\delta_1>0$ such that $0<x<\delta_1\implies 1/x>\delta$. Therefore,
$$0<x<\delta_1\implies\frac{1}{x}>\delta\implies\left|f\left(\frac{1}{x}\right)-L\right|<\varepsilon$$
This shows that
$$\lim\limits_{x\to\infty}f(x)=L\implies \lim\limits_{x\to 0^{+}}f\left(\frac{1}{x}\right)=L$$
Now let's prove the converse. Since $\lim_{x\to 0^{+}}f(1/x)=L$, we have that for all $\varepsilon >0$, there exists a $\delta >0$ such that $0<x<\delta\implies \left|f(1/x)-L\right|<\varepsilon$. Since $\lim_{x\to\infty}1/x=0$, there exists a $\delta_1$ such that $x>\delta_1 \implies\left|1/x-0\right|=1/x<\delta$. Therefore,
$$x>\delta_1\implies\frac{1}{x}<\delta\implies\left|f\left(\frac{1}{\frac{1}{x}}\right)-L\right|=|f(x)-L|<\varepsilon$$
This shows that
$$\lim\limits_{x\to 0^{+}}f\left(\frac{1}{x}\right)=L\implies\lim\limits_{x\to\infty}f(x)=L$$
Thus,
$$\lim\limits_{x\to\infty}f(x)=L\iff\lim\limits_{x\to 0^{+}}f\left(\frac{1}{x}\right)=L$$
Let me know if I made any mistakes!