Is this a legal transformation? To be found: $$\lim \left(1+\frac{2}{n}\right)^n$$
Presuppose $~~\lim \left(1+\frac{1}{n}\right)^n=e~~$ is already shown.
Expanding the first equation: $$\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{2\cdot\frac{n}{2}}=\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}\cdot\lim \left(1+\frac{1}{\frac{n}{2}}\right)^{\frac{n}{2}}$$
I'd say that those two factors converge against $e$, just not as fast. My explanation would be, that the sets $\mathbb N$ and $\{\frac{n}{2}:n\in \mathbb N\}$ have infinitely many elements in common.
So is it enough to just (state that and) add
$$=\lim \left(1+\frac{1}{n}\right)^n\cdot \lim \left(1+\frac{1}{n}\right)^n=e\cdot e=e^2$$?
 A: Your substitution is valid: if $u(x)$ is monotic increasing then $\displaystyle\lim_{x\to\infty} f(x)=\lim_{x\to\infty} f(u(x))$. However as pointed out by Ian, this has nothing to do with a $\Bbb N$ and $u(\Bbb N)$ having an infinite number of elements in common. It is fairly straightforward to find an $f$ and $u$ where $\Bbb N\cap u(\Bbb N)$ is infinite but the limit $\lim f$ does not exist where $\lim f\circ u$ does. Try to find one! (However if the first exists, then the second must also exist and be equal to it. Can you prove this too?)
If $\lim a_n$ and $\lim b_n$ exist (and are finite), then $\lim (a_nb_n)$ also exists and equals $(\lim a_n)(\lim b_n)$.
Alternatively, since the map $x\mapsto x^2$ is continuous, we have $\lim a_n^2=(\lim a_n)^2$, more direct.
All of these facts fall under the heading of real analysis. Are you familiar with these facts? If so, then your argument is good. Otherwise, perhaps familiarize yourself with some basic analysis.
A: Just for your curiosity, the problem can also be approached using Taylor series; for large values of $n$, can be established $$\left(1+\frac{a}{n}\right)^n=e^a-\frac{a^2 e^a}{2 n}+\frac{a^3 (3 a+8) e^a}{24 n^2}-\frac{a^4 (a+2) (a+6) e^a}{48
   n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$
