Effect on the limiting behavior of multiplication of pre-image by a function converging to 1 Let $f: \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ be monotonic (and as smooth as you like), satisfying $\lim_{x \rightarrow \infty} f= \infty$. Let $g: \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ also be monotonic, satisfying $g(x) \leq 1$ everywhere and $\lim_{x \rightarrow \infty} g= 1$. If it helps, we can also assume that $g$ enjoys any amount of smoothness we desire. 
$\textbf{Question}$: Under what hypotheses on $f$ is it the case that for any and all $g$ satisfying the above, one has 
$\lim_{x \rightarrow \infty} \frac{f(g(x)*x)}{f(x)} = 1$ ?  
I believe this should hold whenever $f$ is a polynomial. For example when $f = x^{n}$, the numerator in the above quotient will be $[g(x)]^{n} \cdot x^{n}$, and the first factor there will converge to $1$. Since that computation didn't depend on $n$ being integral, this also works whenever $f$ is a (positive) power function. One can check that $f= \log(x)$ or $f= x*\log(x)$ also works. 
However, if $f= 2^{x}$, the quotient above becomes $2^{[g(x)- 1]*x}$, and note that the exponent there is negative. Hence if $g$ converges to $1$ extremely slowly, this could conceivably converge to $0$. 
So my guess is that whenever $f$ is little o of some polynomial, this should hold, but I'm having trouble coming up with a proof. I'd rather not assume analyticity of $f$ so I don't have Taylor approximations to play with, although if there is a simple proof using the Taylor series and assuming that f is analytic, I would also be interested in seeing that.  
Thank you for your time!
 A: I have a counterexample and a condition that works.
Counterexample
Here's an example where $f$ is little-o of a polynomial but the given result does not hold.  Let
$$
f(x) \,=\, 2^{\lfloor \log_2 x\rfloor}\qquad\text{and}\qquad g(x)=\frac{x-1}{x}.
$$
Note that $f(x)$ is always an integer, namely the greatest power of $2$ that is less than or equal to $x$.  Since $f(x)\leq x$ for all $x$, the function $f(x)$ is $O(x)$.  Then for any integer $n$,
$$
\frac{f\bigl(g(n)\,n\bigr)}{f(n)} \,=\, \frac{f(n-1)}{f(n)} \,=\, \begin{cases}1/2 & \text{if $n$ is a power of $2$} \\[6pt] 1 & \text{otherwise}\end{cases}
$$
Thus
$$
\lim_{x\to\infty} \frac{f\bigl(g(x)\,x\bigr)}{f(x)}
$$
does not exist.
Of course, the function $f$ is not smooth, but there exists a nondecreasing $C^\infty$ function that agrees with $f$ on the integers, so no amount of smoothness will suffice.  We can also make $f$ strictly increasing, if desired, by allowing it to increase by a small amount $\epsilon$ on the interval $[2^n,2^{n+1}-1]$ for each $n$.
Condition that Works
Suppose that $f$ is differentiable and there exists a constant $C$ so that
$$
x\,f'(x) \,\leq\, C\,f(x)
$$
for all sufficiently large $x$. (This holds, for example, if $f(x)$ is any polynomial.) I claim that the statement holds in this case.
To see this, let $F(x) = \log f(x)$.  Then $F$ is differentiable and
$$
x\,F'(x) \,\leq\, C
$$
for all sufficiently large $x$.  Then
$$\begin{multline*}
\left|\log\left(\frac{f\bigl(g(x)\,x\bigr)}{f(x)}\right)\right| \,=\, \bigl|F(g(x)\,x) - F(x)\bigr| \\[6pt]
=\, \int_{g(x)\,x}^{x} F'(t)\,dt \,\leq\, \int_{g(x)\,x}^{x} \frac{C}{t}\,dt \,=\, -C \log g(x)
\end{multline*}$$
for all sufficiently large $x$.  Since $g(x) \to 1$ as $x\to\infty$, it follows that
$$
\lim_{x\to\infty} \frac{f\bigl(g(x)\,x\bigr)}{f(x)} \,=\, 1.
$$
