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.
Best Answer
Let $L=Lim_{x\to a} f(x)^{g(x)}\to 1^{\infty}$ as $\lim_{x \to a}f(x)=1+\epsilon, \lim_{x \to a} g(x)=\infty$, then $$L=\lim_{x\to a} f(x)^{g(x)}= \exp[\lim_{x\to a}g(x)\log f(x)] =\lim_{x\to a} \exp[g(x)\log(1+\epsilon)]$$ Used $\log(1+\epsilon)\approx \epsilon$ and $\epsilon$ is in turn $f(x)-1$. So we have $$L=\exp[\lim_{x \to a} g(x)[f(x)-1]]~~~~(1).$$ Using this here we get $$L=\exp\left[\lim_{x\to \infty} (2x-5)\frac{-4x-1}{x^2+x+2}\right]= e^{-8}.$$