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

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.

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    $\begingroup$ 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? $\endgroup$ – 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.


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