I'd like to calculate the following by the only sequence concept, not to involve the other concepts like limit of function, etc.
problem
$\begin{align} \tag{i} &\lim_{n\rightarrow \infty} {\left({1\over2}+ {1 \over n}\right)}^n \\ \tag{ii} &\lim_{n\rightarrow \infty} {\left(\frac{2n+3}{2n-1}\right)} \end{align} $
Question
The first limit is calculated as followings. $${({1\over2}+ {1 \over n})}^n = {1 \over 2^n}{\{(1+ {1 \over {n/2}})^{n/2}\}^2}$$ When $n$ goes infinity, then $0e^2 =0$, so we have all done.
However how can we verify $\lim (1+ {1 \over {n/2}})^{n/2} = e$? I think it can be only verified by the theorem that any subsequence of a convergent sequence is convergent to the same limit as the original. To be more precisely, let $e_n = (1+ 1/n)^n$, then $\lim e_n = e$ and think about its subsequence $e_{2n}$, then $\lim (1+1/2n)^{2n} = e$.
However here the $e_{n/2}$ cannot be subsequence of $e_n$ because the domain of subsequence should be $N$ and the codomain should be also $N$.
Now the second calculation is same as the first.
$$(\frac{2n+3}{2n-1})^n = \{(1+ \frac{1}{(2n-1)/4})^{(2n-1)/4} \}^{4n/(2n-1)}$$
As the above logic, $e_{(2n-1)/4}$ also cannot be subsequence.
We usually use this kind of solutions when calculating limits, but I don't know why it makes sense.
Those substitutions work because $f(x)=(1+1/x)^x$ is an increasing function for $x>0$. If $f(x)$ is monotone and $\lim\limits_{n\to\infty}f(n)=L$ then for every divergent sequence $a_n$ (that is $\lim\limits_{n\to\infty}a_n=+\infty$) it is straightforward to prove that $\lim\limits_{n\to\infty}f(a_n)=L$.
But if $f(x)$ is not monotone then the above theorem does not hold. Take for instance $a_n=\sin(n\pi)$: you have $\lim\limits_{n\to\infty}a_n=0$ but $\lim\limits_{n\to\infty}a_{n/2}$ does not exist.