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.

$$\lim\limits_{n \to \infty } \left(\frac{n+1}{n-1}\right)^{3n^{2}+1}$$
Note the Bernoulli inequality $(1+x)^m\ge1+mx$. Then:
$$\left(\frac{n+1}{n-1}\right)^{3n^{2}+1}=\left(1+\frac{2}{n-1}\right)^{3n^{2}+1}\ge 1+\frac{2(3n^2+1)}{n-1}\ge1+\frac{(2)(3n^2)}{n}=1+6n\to\infty$$
