<s>[INST] Find limit $\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right) $ Define $a_n$ as:
$$
a_n = \left(\left(1+\frac{1}{n^2}\right)^{n^2}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)\right)^n
$$
Now I want to calculate $\lim_{n \to \infty} a_n$. So, real question is about other limit.
$$
\lim_{n\to\infty}\left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right)^n
$$
As $\lim_{n\to\infty}\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n = 1$, we cannot conclude limit in easy way. ($1^{\infty}$ case.) I don't know, what I should do now.
I tried:
$$
\ln \left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right) =
n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right)
$$
So, now problematic is limit:
$$
\lim_{n \to \infty} n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right)
$$
It's not better. How should I calculate this limit?
 [/INST] Let $f_n$ be the sequence given by $f_n=n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)$.  We can evaluate the limit of $f_n$ by expanding the logarithm functions embedded in the limit of interest as 
$$\begin{align}
\lim_{n\to \infty}f_n&=\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)\\\\
&=\lim_{n \to \infty} n\left(n^2\left(\frac{1}{n^2}+O\left(\frac{1}{n^4}\right)\right)+n\left(-\frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)\right)\\\\
&=-\frac12
\end{align}$$
</s>