I'm have some doubts about how to correctly solve a complex limit (for me!). I know that I should use kind of asymptotics and the hierarchical order of infinite.
This is the limit, and my idea. I checked with an online calculator and the result is correct, but the calculator applies L'Hospital rule but I can't apply it because I can't use derivative.
$$\lim_{n->\infty} \frac{n+\log{n^2}-2^n}{2(\log{n})^3+n\log{n}+n^2}$$
My idea:
1) For the numerator:
$$\lim_{n->\infty} {n+\log{n^2}-2^n} \sim -1*2^n$$
Considering the hierarchy, $-2^n$ goes first to the infinity, so I can just study $-2^n=-1*2^n$
2) For the denominator, I apply the same idea:
$$\lim_{n->\infty} {2(\log{n})^3+n\log{n}+n^2} = 2(\log^3{n})+nlogn+n^2 \sim n^2$$
So I can study:
$$\lim_{n->\infty} \frac{-1*2^n}{n^2}$$
The numerator goes faster than the denominator to the infinity so answer is $-\infty$
I'm not sure about the entire procedure, even if it sounds realistic to me. Sorry if I wrote something horrible! :)
To make it a little more rigourous, you can factor out the dominant terms that youd identified, and
$$\lim_{n\to\infty} \frac{n+\log{n^2}-2^n}{2(\log{n})^3+n\log{n}+n^2}= $$ $$\lim_{n\to\infty} \frac{2^n}{n^2}\frac{n2^{-n}+\log{n^2}2^{-n}-1}{2\dfrac{(\log{n})^3}{n^2}+\dfrac{\log{n}}n+1}.$$
By the ratio test,
$$n^k2^{-n}\text{ and }\log^k\,2^{-n}$$ tend to $0$, as $\left(\dfrac {n+1}n\right)^k2^{-(n+1)+n}$ and $\left(\dfrac{\log(n+1)}{\log n}\right)^k2^{-(n+1)+n}$ tend to $1/2$ for all $k>0$
and this implies, with $m=\log_2n=\log n\log_2e$, that
$$\frac{\log^qn}{n^p}=\log_2^qe\left(m^{q/p}2^{-m}\right)^p$$ also tends to $0$ for $p,q>0$.
Now the limit of the large fraction on the right is $\dfrac{0+0-1}{0+0+1}$ and by the previous statements, the requested limit diverges to infinity.