Evaluate limit of the form $0^{\infty}$ $$\lim_{n\to\infty}\left(\frac{n^4-3\cdot n^3-n^2+2\cdot n-1}{n^5+n^4-n^3-3\cdot n^2-3\cdot n+1}\right)^{\frac{6\cdot n^5-2\cdot n^4-2\cdot n^3+n^2-2\cdot n}{9\cdot n^4-2\cdot n^3+n^2+3\cdot n}}$$
It is $$\lim_{n\to \infty}\left(\frac{1}{n}+o\left(\frac{1}{n^2}\right)\right)^{(n+o(1))}$$
How can I bring it to a form that I can compute?
 A: All limits of the form $0^\infty$ have $0$ as limit. This is not an indeterminate form, like $\infty^0$ or $\frac\infty\infty$.
To have some notation and formality, let $a_n\to 0$ and $b_n\to \infty$ be two sequences, and for simplicity assume that $a_n, b_n> 0$. Then $a_n^{b_n}\to 0$. You can show this, for instance, by comparison:


*

*Option 1: Since $a_n\to 0$ and $b_n\to \infty$, there is some point $N$ where $a_n\leq \frac12$ and $b_n\geq 1$ for all $n\geq N$. Thus we get $$a_n^{b_n}\leq \left(\frac12\right)^{b_n}$$for all $n\geq N$. And the right-hand side certainly goes to $0$.

*Option 2: There is some point $N$ where $a_n\leq 1$ and $b_n\geq 1$ for all $n\geq N$. Then
$$
a_n^{b_n} \leq a_n
$$
for all $n\geq N$. And the right-hand side certainly goes to $0$.


(These arguments work more or less unchanged for negative or zero $a_n, b_n$ as well, as long as $a_n^{b_n}$ makes sense. But I skipped it as I feel it's more difficult to type corrrecly, more difficult to read, and doesn't actually give that much more value to this post.)
A: This is one of the many cases where composition of Taylor series is useful.
We have $y_n=A_n^{B_n}$ where
$$A_n=\frac{n^4-3 n^3-n^2+2 n-1}{n^5+n^4-n^3-3 n^2-3 n+1}\quad \text{and} \quad B_n=\frac{6 n^4-2 n^3-2 n^2+n-2}{9 n^3-2 n^2+n+3}$$
We shall work with
$$\log(y_n)=B_n \log(A_n)$$
Using the long division, we have that
$$A_n=\frac{1}{n}-\frac{4}{n^2}+\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(A_n)=-  \log(n)-\frac{4}{n}-\frac{4}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$B_n=\frac{2 n}{3}-\frac{2}{27}-\frac{76}{243 n}-\frac{377}{2187
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(y_n)=-\frac{2}{3} n \log (n)+\frac{2 }{27}\log (n)-\frac{8}{3}+\frac{4 (19 \log (n)-144)}{243 n}+O\left(\frac{1}{n^2}\right)$$ Now,using the expansion to $O\left(\frac{1}{n}\right)$
$$y_n=e^{\log(y_n)}={e^{-8/3}} n^{2/27}n^{-2 n/3}=n^{\frac{2(1-9 n)}{27} }e^{-8/3}$$ as @Gary wrote in comments.
For $n=100$, the relative error is $0.97$% and for $n=1000$, it is $0.02$%.
Edit
We can make it more general for the case where
$$A_n=\frac{a}{n}+\frac{b}{n^2}+O\left(\frac{1}{n^3}\right)\quad \text{and} \quad B_n=c n+d+O\left(\frac{1}{n}\right)$$ and get
$$y_n=e^{\frac{b c}{a}} \left(\frac{a}{n}\right)^{c n+d}$$
A: There is no hesitation with $0^\infty$, which can be read as $\dfrac1{\infty^\infty}$.
