Finding the limit of the sequence $x(n) = (1+2/n)^n$ What we are allowed to use -
1) The fact that limit of $(1+1/n)^n$ exists and assumed to be some real number $e$
2) Subsequencial properties of limits of sequences
3) Basic properties of limit
In the previous question
$x(n) = (1+1/n^2)^{(2(n^2))}$
We just used the fact that the given sequence is square of the subsequence of the sequence $(1+1/n)^n$ 
And shall thus converge to $e^2$
I'd expect we'd be required to use something similar in this question but I am unable to, can't see a valid subsequence forming.
Edit - This question is not a duplicate of the question suggested as that question assumes that a lot more theorems have been proven, especially, Limits of log.
 A: The limit that defines the constant $e$ is
$$
e=\lim_{n\to\infty}\left(1+\frac1n\right)^n
$$
Since $x^2$ is a continuous function, we get
$$
e^2=\lim_{n\to\infty}\left(1+\frac2n+\frac1{n^2}\right)^n
$$
It is simple to see that
$$
\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n\le\left(1+\frac2n\right)^n\left(1+\frac1{n^2}\right)^n
$$
which is the same as
$$
\left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac1{n^2+1}\right)^n\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n
$$
which implies by Bernoulli's Inequality
$$
\left(1+\frac2n+\frac1{n^2}\right)^n\left(1-\frac{n}{n^2+1}\right)\le\left(1+\frac2n\right)^n\le\left(1+\frac2n+\frac1{n^2}\right)^n
$$
The Squeeze Theorem says
$$
\lim_{n\to\infty}\left(1+\frac2n\right)^n=e^2
$$
A: Let $n=2m$. Then $\lim_{n\to+\infty}x(n)=\lim_{m\to+\infty}x(2m)$ and $$\left(1+\frac{2}{n}\right)^n=\left(1+\frac{2}{2m}\right)^{2m}=\left(\left(1+\frac{1}{m}\right)^{m}\right)^2=\left(1+\frac{1}{m}\right)^{m}\left(1+\frac{1}{m}\right)^{m}$$ 
A: Considering $\displaystyle \sqrt{x(2n)}=\left( 1+\frac{1}{n} \right)^{n}$,
as we know $\displaystyle \lim_{n\to \infty} \left( 1+\frac{1}{n} \right)^{n}=e$,
so $\displaystyle \lim_{n\to \infty} \sqrt{x(2n)}=e$,
$\displaystyle \lim_{n\to \infty} x(2n)=e^{2}$
$\displaystyle \lim_{n\to \infty} \left( 1+\frac{2}{n} \right)^{n}=e^{2}$
A: Consider $$x_n=\Big(1+\frac 2n\Big)^n$$ Take logarithms $$\log(x_n)=n \log\Big(1+\frac 2n\Big)$$ Now, using that, for small $y$, $\log(1+y)\simeq y$, then $$\log(x_n)\simeq n \times \frac 2n=2$$
