Contradictive statement for a limit problem 
Show using the definitions that the sequence $\frac{n^2+1}{n+5}$ does not converge.

If one assumes that $\frac{n^2+1}{n+5}$ converges to say $a$, then $$\left| \frac{n^2+1}{n+5} - a\right| < \varepsilon$$ for any $\varepsilon >0$ when $n \ge n_0$ for some $n_0 \in \Bbb N$.
In particular this should hold when $\varepsilon =1/2$.
I know get that $$ -\frac12 \le \frac{n^2+1}{n+5} -a < \frac12$$
but I don't think I can get a contradicition from here. I was thinking I could contradict on $a$ not being unique, but doesn't seem like that would work. What other approaches are there?
 A: Your approach is fine, you are almost there. Polynomial division gives
$$\frac{n^2+1}{n+5} = n - 5 + \frac{26}{n+5}.$$
Can you now show that this cannot be bounded by $a + \frac{1}{2}$?
A: You could say that $$a_n:=\frac{n^2+1}{n+5}$$ behaves asymptotically like $$\frac{n^2}{n} = n$$
Thus for any $\varepsilon$ you get that for all $n>N$ large where you choose $N>a+\varepsilon$
$$|a_n-a|>|N-a|> \varepsilon$$
which contradicts a convergence (unless you include infinity).
A: By reduction ad absurdum: suppose that the sequence $(a_{n})_{n\in \mathbb{N}}$ defined by $a_{n}:=\frac{n^{2}+1}{n+5}$ converges to  $\ell\in \mathbb{R}$, so by definition
$$\forall \varepsilon>0\,  \exists n_{0}\in \mathbb{N};\, n>n_{0} \implies  |a_{n}-\ell|<\varepsilon.$$
Fixed $\varepsilon=\varepsilon_{0}$, then there exists $n_{0}\in \mathbb{N}$ such that for all $n>n_{0}$ we have $$\left|\frac{n^{2}+1}{n+5}-\ell\right|<\varepsilon_{0} \iff -\varepsilon_{0} +\ell<\frac{n^{2}+1}{n+5}<\varepsilon_{0}+\ell,$$
but then $$\frac{n^{2}+1}{n+5}<\varepsilon_{0}+\ell \iff\frac{26}{n+5}+n<(\varepsilon_{0}+5)+\ell.$$
Contradiction. Hence $(a_{n})_{n\in \mathbb{N}}$ does not converges.
