Deducing whether a limit can be separated into two limits If one knows that $$ \lim_{x->\infty}(f(x)g(x)) \text{ is convergent}$$ and that $$ \lim_{x->\infty}f(x) \text{ is convergent}, $$
one might be tempted to assume that $$ \lim_{x->\infty}g(x) \text{ is also convergent}, $$ but this is not necessarily the case. Just pick $g(x) = x$, $f(x) = \frac{1}{x}$ to see that. In that particular case this is so because $0 * \infty$ is undetermined. However, if the limit of $f(x)$ were to converge to a value other than $0$, is it sufficient, along with the knowledge of the limit-of-product value, to say that the limit of products is the product of the limits? Notice that $0$ is the only value that can "have an impact" on $\infty$.
This is different from checking both the limits of $f(x)$ and $g(x)$ separately, hence my question.
 A: Similar to proving the product rule for limits, it is also possible to prove the corresponding rule for quotients:
If
$$\lim_{x \to l} u(x) = a \text{ and }\lim_{x \to l} v(x) =b \text { and }b \neq 0,$$ then
$$\lim_{x \to l} \frac{u(x)}{v(x)} = \frac{a}b,$$
where $l$ can be any real (or complex) number, including the (informally speaking) "infinite" values $\infty, -\infty$ and $+\infty$.
Your question can thus be answered in the affirmative, setting $u(x)=f(x)g(x), v(x)=f(x).$
A: If $\lim_{x\to \infty} f(x)\neq 0$ then $f(x)\neq 0$ for large $x$ so we have
$$\lim g(x)=\lim\frac{f(x)g(x)}{f(x)}=\frac{\lim f(x)g(x)}{\lim f(x)}.$$
A: Note the limit of $g(x)$ can be written as:
$$\lim_{x\to\infty} g(x)=\lim_{x\to\infty} \frac{f(x)g(x)}{f(x)}$$
If $\lim f(x)=L$ exists and $L\neq 0$, then you can write it as
$$\lim_{x\to\infty} g(x)=\lim_{x\to\infty} \frac{f(x)g(x)}{f(x)}=\frac{\lim_{x\to\infty} f(x)g(x)}{\lim_{x\to\infty} f(x)}$$
A: You have
$$g(x) = \frac{f(x) g(x)}{f(x)}.$$  Hence if $\lim_{x \to \infty} f(x)g(x) = a$ and $\lim_{x \to \infty} f(x) = b \neq 0$ you have $$\lim_{x \to \infty} g(x) = \frac{a}{b}.$$
