Proving a limit converges to -3 using definition of convergence.

Question

so I have the problem $\lim\limits_{n \to oo} \frac{2-3n^2}{n^2+2n+1}$. I have to prove this using the epsilon definition. So I know the limit equals -3. So I do |$\frac{2-3n^2}{n^2+2n+1}$ + 3 | < $\epsilon$. Simplifying this I get to the point, $\frac{5+6n}{(n+1)^2}$ < $\epsilon$ and I don't know how to process from here. Any tips or help would be greatly appreciated! Every other example I found using the definition has cancellation in the numerator to a term with no n in it so since this example has an n in both the numerator and denominator with no way to simplify I am lost.

Answer 1

You already have found that $$\left| \frac{2-3n^2}{n^2+2n+1} + 3 \right|= \frac{5+6n}{(n+1)^2}$$

Now notice that $(n+1)^2 > n^2$ for all $n\in \mathbb{N}$, thus $$\frac{5+6n}{(n+1)^2} < \frac{5+6n}{n^2} $$

Again we notice that for $n\ge 6$, we have that $6n+5<7n$, hence

$$\frac{5+6n}{(n+1)^2} < \frac{5+6n}{n^2} < \frac{7n}{n^2}=\frac{7}{n}$$

Now $\dfrac{7}{n} < \varepsilon$ iff $n>\dfrac{7}{\varepsilon}$. Hence we choose $N=\max \left\{ \dfrac{7}{\varepsilon } , 6 \right\}$. Now if $n \ge N$, then clearly we have

$$\left| \frac{2-3n^2}{n^2+2n+1} + 3 \right| < \varepsilon$$.