For polynomials $p,q$ of the same degree, find $\lim_{x\to+\infty}\left(\frac{p(x)}{q(x)}\right)^x$ without using l'Hôpital's rule

Question

How to use the result $\lim_{x\to+\infty}\left(1+\frac{1}{x}\right)^x=e$ to find $\lim_{x\to+\infty}\left(\frac{x^2-2x-3}{x^2-3x-2}\right)^x$? I rewrite the fraction as $1+\frac{x-1}{x^2-3x-2}$ in the hope of using the result about $e$, but then I have no clues. I wanna know if there's a way that doesn't apply l'Hôpital's rule.

Answer 1

In general $$(1+f(x))^x=\left(1+f(x)\right)^{\frac{xf(x)}{f(x)}}=\left[(1+f(x))^{\frac{1}{f(x)}}\right]^{xf(x)}$$ so if $(1+f(x))^x$ has the indeterminated form $1^{\infty}$ the limit is equal to $e^{\lim_{x\to\infty}xf(x)}$. In this particular case $f(x)=\left(1+\frac{x-1}{x^2-3x-2}\right)$ so $$\lim_{x\to\infty}x\left(1+\frac{x-1}{x^2-3x-2}\right)=1$$

Answer 2

$$\lim_{x\to+\infty}\left(\frac{x^2-2x-3}{x^2-3x-2}\right)^x=\lim_{x\to+\infty}\left(1+\frac{x-1}{x^2-3x-2}\right)^x=\lim_{x\to+\infty}\left(1+\frac{1}{\frac{x^2-3x-2}{x-1}}\right)^x$$

Now

$$\lim_{x\to+\infty}\left(1+\frac{1}{\frac{x^2-3x-2}{x-1}}\right)^x=\lim_{x\to+\infty}\left[\left(1+\frac{1}{\frac{x^2-3x-2}{x-1}}\right)^{\frac{x^2-3x-2}{x-1}}\right]^{\left(\frac{x-1}{x^2-3x-2}\right)\cdot x}=\lim_{x\to+\infty}e^1=e$$