Identity with exponential function: $\lim_{n\to\infty}\frac{n^{2n}}{(n+1)^{2n}} = \frac{1}{e^2}$

Question

Could you please explain me how we got this identity

$\lim_{n\rightarrow \infty}\frac{n^{2n}}{(n+1)^{2n}} = \frac{1}{e^2}$

when we know

$\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n = e$

Thanks!

Answer 1

$$\lim _{ n\rightarrow \infty } \frac { n^{ 2n } }{ (n+1)^{ 2n } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { 1 }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } \right) }^{ 2 } } =\frac { 1 }{ { e }^{ 2 } } $$

Answer 2

$\lim_\limits{n\to \infty}\frac{n^{2n}}{(n+1)^{2n}}$ divide top and bottom by $n^{2n}$

$\lim_\limits{n\to \infty}\frac{1}{(1+\frac 1n)^{2n}} = \frac 1{e^2}$

Answer 3

Hint we can write it as $(\frac{1}{(1+\frac{1}{n})})^{2n}=\frac{1}{e^2}$