multiplication with limits I am wondering if the following can always be said, or if there are any restrictions:
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=a,\,|a|<\infty$$
then its true that:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}a.g(x)$$
My thoughts are that this can be said, but is not relevant if:
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)\to0,-\infty,\infty$$
Thanks in advance.
 A: The limit is $=a$ means that, for great $x$,
$$\frac{f(x)}{g(x)}=a+\epsilon(x)$$
with
$$\lim_{x\to+\infty}\epsilon(x)=0$$
thus
$$f(x)=ag(x)+g(x)\epsilon(x)$$
so
$$\lim_{x\to+\infty}(f(x)-ag(x))$$
depends strongly on the unknown function $x\mapsto \epsilon(x)$.
A: As I pointed it out in the comments, the limits might not exist. But if $\exists \lim\limits_{\infty}g =:b \in \mathbb{R}\setminus\{0\}$, then we have that $\forall \varepsilon >0 \exists x_0$ so that $\forall x>x_0$
$$\left|\frac{f(x)}{g(x)}-a\right|<\varepsilon$$
and
$$|g(x)-b|<\varepsilon$$
Which means that
\begin{align}
\left|f(x)-ab\right|
&=\left|(f(x)-ag(x))+a(g(x)-b)\right|\\
&\leqslant|f(x)-ag(x)|+|a||g(x)-b|\\
&<\varepsilon(|a|+|g(x)|)\\
&<\varepsilon(|a|+|\varepsilon+b|)
\end{align}
So we can make it as small as we want.
A: No, This is trying to will limits into existance.
$\lim 1 = 1$ but $\lim \frac {f(x)}{f(x)} = 1$.  Just let $f(x)$ be a function without a defined limit.
However if you are given that $\lim f(x)$ and $\lim g(x)$ exist then
if $\lim f(x) = m; \lim g(x) = k\ne 0$ then $\lim \frac{f(x)}{g(x)}=\frac mk = a$ so $\lim f(x) = m = ak = \lim a*g(x)$.
If $\lim f(x) = m\ne 0$ or $\lim f(x) =\pm \infty$; while $\lim g(x) = 0$ then $\lim \frac {f(x)}{g(x)}\ne a$.
If $\lim f(x) =0$ and $\lim g(x) = 0$ then $\lim f(x) = \lim a*g(x)$ for all $a$.
If $\lim f(x) = m; \lim g(x) =\pm \infty$ then $\lim\frac {f(x)}{g(x)} = 0 = a$ and $0*g(x) = 0$ so $\lim ag(x)= 0$ which is not necessarily equal to $m$.  
So say  $f(x) =1$ and $g(x) = x$ would be an exception.  $\lim f(x) = 1\ne 0 =\lim 0*g(x)$.
If $\lim f(x) = \infty$ and $\lim g(x)\ne \infty$ then $\lim\frac{f(x)}{g(x)}\ne a$.
$\lim f(x) = \pm \infty$ and $\lim g(x) = \pm \infty$ but $\lim \frac {f(x)}{g(x)} = a\ne 0$ then $\lim f(x) = \pm \infty$ and $\lim a*g=\pm \infty$.  (I'll leave the signs to be self-evident) 
Final case to explore is $\lim f(x) = \pm \infty$ and $\lim g(x) = \pm \infty$ but $\lim \frac {f(x)}{g(x)} = 0$ then $\lim f(x) = \pm \infty$ but $\lim 0*g(x) = 0$.
So there are 3 exceptions:  $\lim f; \lim g$ don't exist; $\lim f\ne 0$ but $\lim g(x)=\pm \infty$; or $\lim f(x)=\pm \infty; \lim g(x) = \pm \infty$ but $\lim\frac {f(x)}{g(x)} = 0$.
A: This is not true as pointed out in the comments because the limits of $f,g$ may not exist at $\infty$. Although, the statement
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=a,\,|a|<\infty$$
is equivalent to the statement
$$f(x)\in\Theta(g(x))$$
where $\Theta$ denotes Big Theta notation.
