How to calculate $\lim\limits_{n \to \infty } \left(\frac{n+1}{n-1}\right)^{3n^{2}+1}$? I have $\lim_{n \to \infty } \biggr(\dfrac{n+1}{n-1}\biggr)^{3n^{2}+1}$
and I use this way: $\lim_{n \to \infty } \frac{n+1}{n-1}=|\frac{\infty}{\infty}|=\lim_{n \to \infty } \frac{1+\frac{1}{n}}{1-\frac{1}{n}}=1$ and $\lim_{n \to \infty } 3n^{2}+1=\infty$
Then $\lim_{n \to \infty } \frac{n+1}{n-1}^{3n^{2}+1}=|1^{\infty}|$
Continue using the formula $\lim_{n \to \infty } (1+\frac{1}{n})^{n}=e$, I decide to limit and get an answer: $\infty$, and it is right.
But for some reason I can imagine $\lim_{n \to \infty } \frac{n+1}{n-1}^{3n^{2}+1}$ how $\lim_{n \to \infty } \frac{n+1}{n-1}=|\frac{\infty}{\infty}|=\lim_{n \to \infty } \frac{1+\frac{1}{n}}{1-\frac{1}{n}}=1$ and $\lim_{n \to \infty } 3n^{2}+1=\infty$ ?
I can't find such a property of limits.

Hint 
Consider $$a_n=\left( \frac{n+1}{n-1}\right)^{3n^{2}+1}$$ $$\log(a_n)=(3n^2+1)\log\left( \frac{n+1}{n-1}\right)=(3n^2+1)\log\left( \frac{1+\frac 1n}{1-\frac1n}\right)$$ Now, use Taylor for infinitely large $n$ $$\log\left( \frac{1+\frac 1n}{1-\frac1n}\right)=\frac{2}{n}+\frac{2}{3 n^3}+O\left(\frac{1}{n^5}\right)$$
