Without using logarithm rules, how to calculate the limit $\lim_{n\to\infty} R(n)^n$ for a rational function $R$? Well, without using logarithm rules, how to calculate this limit? $$\lim _{n\to \infty }\left(\frac{n^2+8n\:-1}{n^2-4n-5}\right)^n$$  
I can't find any more "nice" presentation of this fraction, and i can't use $\ln $ rules..

$$\frac{n^2+8n-1}{n^2-4n-5}=1+\frac{n^2+8n-1}{n^2-4n-5}-1=1+\frac{12n+4}{n^2-4n-5}$$
$$\left(\frac{n^2+8n-1}{n^2-4n-5}\right)^n=\left[\left(1+\frac{12n+4}{n^2-4n-5}\right)^\dfrac{n^2-4n-5}{12n+4}\right]^{\dfrac{n(12n+4)}{n^2-4n-5}}$$
For $\displaystyle\lim_{n\to\infty}\left(1+\frac{12n+4}{n^2-4n-5}\right)^\dfrac{n^2-4n-5}{12n+4}=e$
For the exponent,
$\displaystyle\lim_{n\to\infty}\dfrac{n(12n+4)}{n^2-4n-5}=\lim_{n\to\infty}\dfrac{12+4/n}{1-4/n-5/n^2}=12$
