Is there any general rule for finidng limits like this:
$\lim_{n\to \infty} \ n^a\ln(\frac{W_b(n)}{W_c(n)})$
where $W_b(n), W_c(n)$ are the polynomials of $b,c$ degree, respectively, without using de l'Hospital's rule?
For example:
$\lim_{n\to \infty} \ n\ln(\frac{n^2+n+6}{n^2-n+12})$
Write $$\frac{W_b(n)}{W_c(n)} = Rn^{b-c}\frac{1 + \dfrac Bn + O(n^{-2})}{1 + \dfrac Cn + O(n^{-2})}$$
Note that $R > 0$, or else the logarithm will not be defined for large $n$.
[If $a \ge 2, b = c, R = 1, B = C$, then the limit will depend on the $O(n^{-2})$ terms, and we keep getting deeper into the rabbit hole.]
Recalling that for small $x$, $$\frac 1{1+x} = 1 - x + O(x^2)\\\ln(1+x) = x + O(x^2)$$ gives $$n^a\ln \frac{W_b(n)}{W_c(n)} = n^a\left((b-c)\ln n + \ln R + (B - C)\frac 1n + O(n^{-2})\right)$$
Also recalling that $n^a\ln n \to 0$ as $n \to \infty$ for all $a < 0$, this means the overall limit for $a < 0$ is always $0$.
If $a \ge 0, b \ne c$, rewrite the expression as $$n^a\ln n\left(b-c + \frac {\ln R}{\ln n} + \frac{B - C}{n\ln n}\right)$$ The last two terms go to $0$ leaving only $b-c$, while $n^\alpha \ln n \to +\infty$. Hence the limit is $\text{sign}(b-c)\infty$.
If $a > 0, b = c, R \ne 1$, there is no logarithm term and $\frac 1n \to 0$. The limit becomes $\text{sign}(\ln R)\infty$.
If $b = c, R = 1$, the expression becomes $$n^{a-1}(B - C) + O(n^{-2})$$
So if $a > 1$, the limit is $\text{sign}(B-C)\infty$, if $a < 1$, the limit is $0$, and if $a = 1$, the limit is $B - C$
If $a = 0, b = c$, the expression is of the form $\ln R + (B - C)\frac 1n + O(n^{-2}$ and the limit is $\ln R$.