Limit of Function raised to a power of another function I'm trying to evaluate the following limit:

$$A =\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}$$

So far, I have exponentiated the limit and whatnot and now I am at this stage:

$$A =\exp\left(\lim_{x\to\infty}(2x-5)\ln\frac{x^2-3x+1}{x^2+x+2}\right)$$

Now, I don't know what to do here, I know via simple algebra that:

$$\lim_{x\to\infty}\ln\frac{x^2-3x+1}{x^2+x+2}=0$$

But $2x-5$ has no limit, and thus I cannot separate $A$ into the product of two limits. Perhaps I am missing something? WolframAlpha says

$$\lim_{x\to\infty}(2x-5)\ln\frac{x^2-3x+1}{x^2+x+2}=-8$$

And that therefore $A = e^{-8}$, but gives no insight as to how this is the case.

One of favorite way to see this so quickly as follows
$$\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}= \lim_{x\to\infty}\left(1-\frac{4x+2}{x^2+x+2}\right)^{2x-5}=\lim_{x\to\infty}\left(1-\frac{1}{\frac{x^2+x+2}{4x+2}}\right)^{\frac{x^2+x+2}{4x+2}\frac{4x+2}{x^2+x+2} (2x-5)} $$
So, since $\lim_{x\to\infty}\frac{(4x+2)(2x-5)}{x^2+x+2}=8.$ Then,
$$\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}= e^{-8}$$
