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.

As I use to say (joke) "we are always closer to $0$ then to $\infty$"
So, start making $x=\frac 1 y$
$$A=\left(\frac{x^2-3 x+1}{x^2+x+2}\right)^{2 x-5}\implies A=\left(\frac{1-3 y+y^2}{1+y+2 y^2}\right)^{\frac{2}{y}-5}$$
Take the logarihm
$$\log(A)=\left(\frac{2}{y}-5\right)\Big[\log(1-3 y+y^2)-\log(1+y+2 y^2)\Big]$$
Work the logarithms separately ... and just finish $\log(A)$.
Continue with Taylor using
$$A=e^{\log(A)}$$ and you are done.
