A question about limits at infinity I got this question:
Let $f, g$ be functions defined on the interval $[0,\infty)$ and let $L \in \mathbb{R}$ be a real number such that $$\lim_{x \to \infty}\left(f(x)\cdot g(x)\right)= L.$$
(1) If $\lim_{x \to \infty}f(x) = m$ where $0 \neq m \in \mathbb{R}$ is a nonzero real number,
Must it be the case that $\lim_{x \to \infty}g(x)$ exist?
(2) If $\lim_{x \to \infty}f(x) = \infty$
Must it be the case that $\lim_{x \to \infty}g(x)$ exist?
I having hard time trying to prove it and to find a counter example. Thanks.
 A: We know that
if $\lim\limits_{x\rightarrow a}{y(x)}=a$ and $\lim\limits_{x\rightarrow a}{w(x)}=b$ then $\lim\limits_{x\rightarrow a}{\left(y(x)\times w(x)\right)}=ab$
Now $\lim\limits_{x\rightarrow \infty}{\dfrac1{f(x)}}=m$ (as $m$ is non-zero)
Consider functions $y(x)=\dfrac{1}{f(x)}$ and $w(x)=f(x)\times g(x)$. Now as both the limits of $y(x)$ and $w(x)$ exists. If you consider the limit of $y(x)\times w(x)\color{grey}{=g(x)}$, you can prove that $g(x)$ approaches the limit $\dfrac{L}{m}$.

About the second question:


*

*As you might have noticed, before defining the limit, we are imposing a condition that $L$ is real I want to avoid things like $\lim{f(x)}=\infty$ and claiming that $\lim f(x)$ exist .We write $f(x)\rightarrow +\infty$ as $x\rightarrow a$ instead of the $\lim$ equals $\infty$). Now this becomes tricky! Although it looks like  the previous proof
works for this one too, it is not so. It cannot be used as the
product of the limits equals the limit of product is applicable only
if the limit of both the functions 'exist', the existence of limit of
$f(x)$ is a problem here, so we have to resolve to more elementary
technique.


We say that a function $f(x)$ approaches $\infty$ if for every $M\in \mathbb{R}$ we can find an $N\in \mathbb{R}$ such that $$x\ge N \implies f(x)\ge M.$$
Now as $f(x)g(x)$ approaches $L$ we can say that for every $\varepsilon^{'} =\varepsilon \cdot M -L$ there exists an $N_1$ such that $x>N_1 \implies -\varepsilon<\frac{L-\varepsilon^{'}}{f(x)}<g(x)<\frac{L+\varepsilon^{'}}{f(x)} <\varepsilon \tag{1}$
And thus $g(x)$ approaches $0$
(For proving (1) you will have to assume that $L\ge0$, for the case $L<0$ you can consider $\varepsilon^{'}=\varepsilon \cdot M +L$ and arrive at the same inequality. Also division by $f(x)$ is valid as we can have an $N_2$ such that for all $x>N_2$, $f(x)$ is greater than $0$)
A: $$
\lim_{x\to\infty}g(x)=\lim_{x\to\infty}{f(x)g(x)\over f(x)}=
{\lim_{x\to\infty}f(x)g(x)\over\lim_{x\to\infty}f(x)}={L\over m}.
$$
(why we can divide by $f(x)$?)
If (2), then
$$
\lim_{x\to\infty}g(x)=\lim_{x\to\infty}{f(x)g(x)\over f(x)}=0
$$
because (bounded thing)/(thing$\to\infty$)$\to 0$.
