Concerning the legality of calculating a limit Question: find the following limit: $$ \lim \limits_{n \to \infty} (1+\frac{1}{2n})^n $$
My approach:
$$ \lim \limits_{n \to \infty} (1+\frac{1}{2n})^n =
\sqrt{\bigg(\lim \limits_{n \to \infty} (1+\frac{1}{2n})^n\bigg)^2} =
\sqrt{\lim \limits_{n \to \infty} (1+\frac{1}{2n})^{2n}} = \sqrt{e} $$
Is the last operation legal? Since we're dealing with infinities here, I'm not sure if I can simply "jump" straight to $ \sqrt{e} $. I'm assuming the inequality $ (\ldots \leq \sqrt{e}) $ is perfectly reasonable.
 A: What you did is not "illegal" but just causes uneasiness. Instead, write:
$$\lim_{n\to \infty} \left( 1 + \frac{1/2}{n}\right)^n$$
which is $e^{1/2}$.
A: It is "legal" because $x\mapsto x^2$ is continuous and the limit you want is finite, but you don't know that beforehand. You could first argument that $$f(n):=\left(1+\frac1{2n}\right)^n \leq \left(1+\frac1n\right)^n,\quad\forall n \in \mathbb{N}$$ and then, taking the limit on both sides, you know that the limit of $f$ is finite and then proceed with what you have done.
A: What you have done is almost correct (you just need to put limit operation before square root symbol and not inside it). The validity of your approach rests on the following result (which is easy to prove via definitions of limit and continuity): 

If $s_{n} \to L$ as $n \to \infty$ and $f$ is continuous at $L$ then $f(s_{n}) \to f(L)$ as $n \to \infty$.

Here we have $s_{n} = (1 + (1/2n))^{2n}$ and $f(x) = \sqrt{x}$. The sequence $s_{n} \to e$ and clearly $f$ is continuous at $e$ and therefore the limit of the desired sequence $f(s_{n}) = (1 + (1/2n))^{n}$ is $f(e) = \sqrt{e}$.
