I am trying to show that $\lim_{x \rightarrow \infty} \frac {P(x)} {{Q}(\log x)} = \infty$, where $\lim_{x \rightarrow \infty} f(x) = \infty \iff \forall N \in \mathbb{R} \exists x_n \in \mathbb{R}^{+} \forall x \in \mathbb{R}^{+} x>x_n \implies f(x)> N$; $P(x) = \sum_{i=0}^{n}c_i \cdot x^{r_i}$ where $n$ is any integer, the $c_i$ are any reals, with the "leading coefficient" (the coefficient of the term of the greatest power) being positive, and the $r_i$ are any non-zero rationals with at least $1$$r_i$ being positive; $Q(x) = \sum_{i=0}^{m}c'_i \cdot x^{r'_i}$ where $m$ is any integer, the $c'_i$ are any reals, with the "leading coefficient" being positive, and the $r'_i$ are any non-zero rationals with at least one $r' _ i$ being positive; and $P(x)$ and $Q(x)$ are functions defined over the positive reals.
Here is my attempt at proving this:
Consider $\hat{Q}(x) = m \cdot C \cdot x^R$ where $C$ is the greatest $c'_i$ and $R$ is any positive integer greater than any of the $r'_i$. Performing the substitution $y= \log x$, $\lim_{x \rightarrow \infty} \frac {P(x)} {\hat{Q}(\log x)} = \lim_{y \rightarrow \infty} \frac {P(e^y)} {\hat{Q}(y)}$; after applying L'hopital's $R$ times, we get that this is $\lim_{y \rightarrow \infty} \frac {\sum_{i=0}^{n}c_i \cdot r_i^R \cdot e^{y(r_i)} }{R!\cdot m \cdot C} = \infty$. $\hat{Q}(x)$ is greater than $Q(x)$ for all $x$, and because $Q(x)$ is positive for all $x$, $\frac {P(x)}{Q(\log x)}$ is greater than $\frac {P(x)} {\hat{Q}(\log x)}$ for all $x$ and diverges to infinity because for any $N \in \mathbb{R}$, we knew that for all values after a certain point $\frac {P(x)} {\hat{Q}(\log x)}>N$; for all values after that same point $\frac {P(x)} {Q(\log x)} > N$.
For the sake of depth, my proof that $\lim_{y \rightarrow \infty} \frac {\sum_{i=0}^{n}c_i \cdot r_i^R \cdot e^{y(r_i)} }{R!\cdot m \cdot C} = \infty$ (which was kind of taken for granted above ) is to show $\lim_{y \rightarrow \infty}{\sum_{i=0}^{n}c_i \cdot r_i^R \cdot e^{y(r_i)} }= \infty$:
we know that if we let $S$ be the term in this sum which has the greatest exponent (and consequently a positive coefficient $c_i$ and exponent $y(r_i)$), and $OS$ be $\sum_{i=0}^{n}c_i \cdot r_i^R \cdot e^{y(r_i)} - S $, $\lim_{x \rightarrow \infty} \frac {OS} S = 0$, as in distributing $1 \over S$ to each of the terms in $OS$, each of the powers of the terms of $OS$ are less than the exponent of $S$, and because $S$'s exponent is positive, each term of $OS$ divided by $S$ becomes the ratio of their coefficients divided by $e^{y (c)}$ for some positive number $c$ (the exponent of the term of $OS$ was either negative, in which case subtracting this from the positive exponent of $S$ in the denominator results in another positive exponent; or it is positive of lesser magnitude, in which case $S$ would also have a positive exponent). As a result of this, we know that there exists a real $n<0$ s.t. for all $x$ after a certain point, $\frac {OS} S > n$. Consequently, for all $x$ after a certain point $S + OS> (1+n)S$, as $OS > n S$, so $S+ OS > S+ n S$. $\lim_{y \rightarrow \infty}(1+n)S = \infty$; so $\lim_{y \rightarrow \infty}S+ OS = \infty$.
Is this proof correct? I was unsure of my overall evaluation of this limit, especially in how I looked at another limit, the one with $\hat{Q}(x)$.