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.
 A: To apply your formula, notice that
$$\left(\frac{n+1}{n-1}\right)^{3n^2+1}=\left(1+\frac2{n-1}\right)^{3n^2+1}$$
Let $u=\frac{n-1}2$ or $n=2u+1$,
$$=\left(1+\frac1u\right)^{12u^2+12u+4}=\left(\left(1+\frac1u\right)^u\right)^{12(u+1)}\left(1+\frac1u\right)^4$$
$$\to e^{12(u+1)}\to e^\infty=\infty$$
Thus, the limit is infinite.
A: $$\lim\limits_{n \to \infty } \left(\frac{n+1}{n-1}\right)^{3n^{2}+1}$$
Note the Bernoulli inequality $(1+x)^m\ge1+mx$. Then:
$$\left(\frac{n+1}{n-1}\right)^{3n^{2}+1}=\left(1+\frac{2}{n-1}\right)^{3n^{2}+1}\ge 1+\frac{2(3n^2+1)}{n-1}\ge1+\frac{(2)(3n^2)}{n}=1+6n\to\infty$$
A: 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 $n$ $$\log\left( \frac{1+\frac 1n}{1-\frac1n}\right)=\frac{2}{n}+\frac{2}{3 n^3}+O\left(\frac{1}{n^5}\right)$$
A: Lots of the confusion in OP seems to due to the careless use of the symbol $\infty$. 
The symbol $\infty$ is not a real number. One should not do the arithmetic carelessly with it. For instance one should have
$$
\lim_{n\to \infty}\frac{2n}{n}=2,
\quad
\lim_{n\to \infty}\frac{3n}{n}=3.
$$
But if one carelessly use the symbol $\infty$, one would have
$$
\lim_{n\to \infty}\frac{2n}{n}=\lim_{n\to \infty}\frac{3n}{n}=\frac{\infty}{\infty}=1
$$
which is nonsense. 

Now go back to the calculation

$$\lim\limits_{n \to \infty } \left(\frac{n+1}{n-1}\right)^{3n^{2}+1}\tag{1}$$

Do you know 
$$
\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e?\tag{2}
$$
(1) and (2) are similar:
$$
\left(\frac{n+1}{n-1}\right)^{3n^{2}+1}=\biggr[\biggr(1+\frac{1}{\frac{n-1}{2}}\biggr)^{\frac{n-1}{2}}\biggr]^{\dfrac{2(3n^2+1)}{n-1}}
$$
A: Take logarithms. Then by the Mean Value Theorem there exists $a\in (n-1,n+1)$:
$\displaystyle{\frac{1}{n+1}<\frac{1}{a}=\frac{\ln (n+1)-\ln (n-1)}{(n+1)-(n-1)}=\frac{\ln (n+1)-\ln (n-1)}{2}}$
Thus 
$\displaystyle{(3n^2+1)\ln (\frac{n+1}{n-1})>\frac{6n^2+2}{n+1}}$
Since $\displaystyle{\frac{6n^2+2}{n+1}\rightarrow +\infty}$ we conclude that $\displaystyle{(3n^2+1)\ln (\frac{n+1}{n-1})\rightarrow +\infty}$ too. And your problem is solved.
