$$\lim\limits_{n\to\infty} \frac{2 − n}{1 + n^2 + n}= 0$$

I know how to generally do these proofs but I'm stuck on this one. I started with:

Given $$\epsilon > 0$$, we want $$N^* \in \mathbb N$$ with the property that $$n \geq N^*$$ implies $$|a_n − a| = \left|\frac{2 − n}{1 + n^2 + n} - 0\right| < ε$$

However, how do I take that term out of absolute value without the values $$n=0,1$$ being omitted? That's my only problem. I can do everything from there.

• You are asking about $n=0, n=1$ in a claim about what happens when $n$ grows large. Can't we select $N>100$ so that we don't have to deal with these cases? – Mason Feb 27 at 20:46

You have indicated that you can handle all the cases except $$n=0$$ or $$n=1$$ so I will assume that you have for any $$\epsilon>0$$ some $$N^*$$ such that for all $$n>N^*>1$$ we have $$|a_n-a|<\epsilon$$. In this case I propose we select $$N^\triangle= \max(N^*,2)$$ and then we have eliminated the annoying cases that bothered us.