If $f(x) \to 0$ and $g$ is a bounded function, then $f(x)g(x) \to 0$ I am using a non-English source text so I am not sure that all technical terms is given their correct English name.
What my source text calls "upper limited function" is defined as a function that has an upper limited range, that is, there is a B such that
$\forall x \in D_{f}: f(x) \leq B$
Question 1: What is the correct English name for this term?
I came across the following theorem:
If $\lim f(x) = 0$ and the function $g(x)$ is limited (i.e. both "upper limited" and "lower limited"), then $f(x)g(x) \rightarrow 0$
Question 2: Is there a specific name for this theorem?
The proof of this theorem starts by stating that the requirement that g(x) is limited for large x implies that there exists numbers $C$ and $\omega_{0}$, such that:
$$x > \omega_{0} \rightarrow | g(x)| < C $$
Then they define $ \epsilon $ as a positive number and the assumptions behind $f(x)$ means that there is a number $\omega_{1}$, such that
$$x > \omega_{1} \rightarrow | f(x)| < \frac{\epsilon}{C} $$
Now, if $\omega = \max(\omega_{0},\omega_{1})$, then
$$x > \omega \rightarrow |f(x)g(x)| = |f(x)||g(x)| < \frac{\epsilon}{C} \cdot C = \epsilon$$
This apparently means exactly that $f(x)g(x) \rightarrow 0$ when $x \rightarrow \infty$
Question 3: I do not really understand that much of this proof, such as the the part about assumptions behind $f(x)$ implies the things it does or how the part about $\omega = \max(\omega_{0},\omega_{1})$ follows. Any tips?
Question 4: I sometimes find myself dealing with non-English source texts for a variety of reasons. Any advice on how to connect knowledge gained from these texts to the larger knowledge amassed from English literature?
 A: I assume you mean $ \lim\limits_{x \rightarrow a} f(x) = 0$ and $ \lim\limits_{x \rightarrow a} f(x)g(x) = 0 $ and let $ a = \infty $.
Ad Question 3: To me this seems to be a basic proof of a convergence via the $ \varepsilon $-criterion.
Let $ \overline{\varepsilon} \gt 0 $.
Make sure you know the definition of convergence regarding a function.
Limited in the context of a function $ g(x) $ means, that there exists a constant C so that for every $ \varepsilon \gt 0 $:
$$ \exists \overline x_g : \forall x > \overline x_g, x \in \mathbb{D}_{g(x)}  : |g(x) - C| \lt \varepsilon $$
$ \mathbb{D}_{g(x)} $ means the domain of $ g(x) $, e.g. the domain of $ \ln(x) $ for real values $ x $ is normally $ \mathbb{R}^{+} \setminus \{ 0 \} $.
We know that $ \lim\limits_{x \rightarrow a} f(x) = 0 $, which implies that for every $\varepsilon$ we can find a $ \overline x_f \in \mathbb{D}_{f(x)} $ so that:
$$ \forall x \gt \overline x_f, x \in \mathbb{D}_{f(x)}:  |f(x)| < \varepsilon $$
As with all proofs where you can find an $ \varepsilon $ you now have to tweak the $ \varepsilon $ to fit your needs.
$ \varepsilon = \frac{\overline{\varepsilon}}{C} $ will do it.
You want to make sure that starting from $ \overline x $ every other $ x \gt \overline x $ satisfies $ |f(x)g(x)| \lt \overline\varepsilon $.
So you must pick the Maximum of the $ \overline x_f $ from $ f(x) $ and the $ \overline x_g $ from $ g(x) $ because for $ x \gt \max(\overline x_f, \overline x_g) $ both implications will be satisfied and we can derive the last inequality I marked with (*).
$$ \forall x \gt \max(\overline x_f, \overline x_g): |f(x)g(x)| = |f(x)| |g(x)| \overset{(*)}{\lt} \frac{\overline{\varepsilon}}{C} \cdot C = \overline{\varepsilon} $$
Therefore: $ \lim\limits_{x \rightarrow a} f(x)g(x) = 0 $.
Ad Question 4: I'm a german and find myself reading english source texts for my studies very often. After a short time I learned to deal with the correspondences of math in german and math in english. (Maybe it is easier for a german because many things in analysis or algebra are almost the same in english and german - eigenvalues, Galois-Theory, many definitions in measure theory, etc.)
Just always try to make the connection between the things you read in your native tongue and how you name them in english.
