I'm trying to prove that this sequence $$\frac{n+2}{2n+3} $$ converges to $1/2$. This is what I have so, far.
Let $\epsilon > 0$. Choose $N=\dfrac{1}{4\epsilon}$ and let $n > N$ where $n$ is some integer. Then $n>\dfrac{1}{4\epsilon}> \dfrac{1}{4\epsilon} -\dfrac{3}{2}$.
So, $4n >\dfrac{1}{\epsilon}-6$ and $4n+6 >\dfrac{1}{\epsilon}$. Thus, $\dfrac{1}{4n+6} < \epsilon$ and hence
$$\left|\frac{n+2}{2n+3}-\frac{1}{2}\right| = \left|\frac{2(n+2)}{2(2n+3)} -\frac{(2n+3)}{2(2n+3)} \right| = \left|\frac{1}{4n+6}\right|=\frac{1}{4n+6} < \epsilon $$
Is there any flaw in my proof?