Proof that a given value is NOT the limit of a sequence. In the sequence: 
$$\lim _{n\rightarrow \infty }{\frac {n+1}{2\,n+3}}\neq 3$$
I know how to prove that the limit is actually $1/2$, but is there another way to prove that 1 is NOT the limit? 
I tried to prove by negation showing that if 1 is the limit I can't find an $N$ that for every epsilon etc etc but got a bit lost. I know this is trivial but would appreciate the help. 
Thanks
 A: For example, show the limit is $\le 1/2$ like this:
$$
\frac{n+1}{2n+3} < \frac{n+1}{2n+2} = \frac{1}{2}
$$
A: Yes, the value of the fraction is always below 1/2. So it stays far away from 3, indeed never gets within 5/2 of it. Note that it is not necessary to know that there is any limit to do answer your question. Anyway, take $\varepsilon = 2$ and no $n$ ever gets you within  $\varepsilon$ of 3.
A: 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 $\frac{1}{2}$.
A: Let $\epsilon=1/4$. Let $N$ be given. Then 
$${n+1\over 2n+3}\le {n \over 2n}\le{1\over2}\quad\Rightarrow|{n+1\over 2n+3} -1|\ge 1/2.$$
This is true for all $n\ge N$. 
So, there is no $N$ such that $|{n+1\over 2n+3} - 1|<\epsilon$
for all $n\ge N$. This shows the sequence does not converge to 1.
