Let $X$ be some number other than $\frac{1}{2}$. Let's define $d=|X-\frac{1}{2}|$, and because $X\neq\frac{1}{2}$ we have $d>0$.
The reverse triangle inequality says that for any $a$ and $b$, $|a-b|\geq ||a|-|b||.$ In particular, $\left|X-\frac{n+1}{2n+3}\right|=\left|X-\left(\frac{n+\frac{3}{2}}{2n+3}-\frac{\frac{1}{2}}{2n+3}\right)\right|=\left|\left(X-\frac{1}{2}\right)-\left(-\frac{1}{4n+6}\right)\right|\geq d-\frac{1}{4n+6}$
Suppose that $\lim_{n\to\infty}\frac{n+1}{2n+3}=X,$ i.e. for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that: for all $n>N$, $\left|X-\frac{n+1}{2n+3}\right|<\epsilon.$ Then we have that for all $n>N$, $d-\frac{1}{4n+6}<\epsilon,$ or equivalently$d<\epsilon+\frac{1}{4n+6}.$ But this is true for all $n>N$ iff $d\leq \epsilon$. And $d\leq \epsilon$ for all $\epsilon>0$ iff $d=0$. But $d>0$; contradiction.
Thus, the limit cannot be anything other than $\frac{1}{2}$.