proof that $ \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1 $ is an equivalence relation Define the following relation on $ \{f \in \mathbb{R}^\mathbb{R} \mid 0 \notin Image(f) \} $:
$ f \ $ is equivalent to $ \ g \ $  if and only if $ \displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1 $.
I want to prove that this is an equivalence relation, is the following proof valid?
$ \underline{\text{Reflexivity:}} $
$ \forall x \in \mathbb{R}, \ \frac{f(x)}{f(x)} = 1 \Rightarrow \displaystyle \lim_{x \to \infty} \frac{f(x)}{f(x)} = \lim_{x \to \infty} 1 = 1 \Rightarrow f \ $ is equivalent to $ f $.
$ \underline{\text{Symmetry:}} $
Assume $ \displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1 $. Therefore, $ \displaystyle \lim_{x \to \infty} \frac{g(x)}{f(x)} = \lim_{x \to \infty} \frac{1}{\frac{f(x)}{g(x)}} = \frac{\displaystyle \lim_{x \to \infty} 1}{\displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)}} = \frac{1}{1} = 1$ and therefore $ \ g \ $ is equivalent to $ \ f $.
$ \underline{\text{Transitivity:}} $
Assume $ \displaystyle \lim_{x \to \infty} \frac{f(x)}{g(x)} = 1 \land \lim_{x \to \infty} \frac{g(x)}{h(x)} = 1 $. Therefore, $ \displaystyle \lim_{x \to \infty} \frac{f(x)}{h(x)} = \lim_{x \to \infty} \frac{f(x)}{g(x)} \frac{g(x)}{h(x)} = \lim_{x \to \infty} \frac{f(x)}{g(x)} \cdot \lim_{x \to \infty} \frac{g(x)}{h(x)} = 1 \cdot 1 = 1 $.
$$\tag*{$\blacksquare$}$$
I feel that in in "Transitivity", I am assuming the limit exists. Is that true?
 A: The first two equals signs are conditional on the remaining limits in the chain existing. But since they do, the steps are valid.
This kind of deferral is pretty standard in limit calculations. But if you want to be more careful, you can organize it this way:

*

*Since $f \sim g$ and $g \sim h$, $\lim_{x\to\infty}\frac{f(x)}{g(x)} = 1$ and $\lim_{x\to\infty}\frac{g(x)}{h(x)} = 1$

*By the product rule for limits, $$ \lim_{x\to\infty}\frac{f(x)}{g(x)}\cdot \frac{g(x)}{h(x)} = \lim_{x\to\infty}\frac{f(x)}{g(x)}\cdot \lim_{x\to\infty}\frac{g(x)}{h(x)} = 1\cdot 1 = 1$$

*Since $\frac{f(x)}{g(x)}\cdot \frac{g(x)}{h(x)} = \frac{f(x)}{h(x)}$,
$\lim_{x\to\infty}\frac{f(x)}{h(x)} = 1$ too.

Therefore, $f \sim h$.
A: The proof is correct. We do not need to assume $\lim \frac{f}{h}$ exists when we prove transitivity. By properties of limit, as you have shown, it follows that
$$1= \lim \frac{f}{g} \lim \frac{g}{h} = \lim \frac{fg}{gh} = \lim \frac{f}{h}. $$
So, $f\sim g\sim h$ does imply that $f\sim h$.
