Prove by definition of a limit

Question

I have this example here:

$\lim_{n \to \infty} \frac{n^{2}+1}{n^{2}+n+1} = 1$

so I have to prove that, $\forall \epsilon > 0, \exists n_{0}\in N, \forall n(n \in N) \geq n_{0} \Rightarrow |\frac{n^{2}+1}{n^{2}+n+1} - 1| < \epsilon$

I have done in these steps

$|\frac{n^{2}+1}{n^{2}+n+1} - 1| < \epsilon$

$|\frac{-n}{n^{2}+n+1}| < \epsilon$

$\frac{n}{n^{2}+n+1} < \epsilon$

$n^{2} + n(1-\frac{1}{\epsilon}) + 1 > 0$

D = $(1-\frac{1}{\epsilon})^{2} - 4 = (1 - \frac{1}{\epsilon} - 2)(1 - \frac{1}{\epsilon} + 2) = -(1 + \frac{1}{\epsilon})(1-\frac{1}{\epsilon}) = \frac{1}{\epsilon^{2}} - 1$

$n_{1,2} = \frac{\frac{1}{\epsilon} - 1 \pm \sqrt{D}}{2}$

I guess we need that

$n > \frac{\frac{1}{\epsilon} - 1 + \sqrt{D}}{2} = \frac{\frac{1}{\epsilon} - 1 + \sqrt{\frac{1}{\epsilon^{2}}- 1}}{2} = \frac{\frac{1}{\epsilon} - 1 + \sqrt{1 + (\frac{1}{\epsilon^{2}}- 2)}}{2} > \frac{\frac{1}{\epsilon} - 1 + \frac{1}{2}(\frac{1}{\epsilon^{2}}- 2)}{2} = \frac{\frac{1}{\epsilon} - 1 + \frac{1}{2\epsilon^{2}}- 1}{2} = \frac{\frac{1}{2\epsilon^{2}} +\frac{1}{\epsilon} -2}{2}$

But I think I have done some mistakes and this last thing could be in more simple form

P.S: How can I verify whether my result $n(\epsilon)$ is true in the end?

Answer 1

Sometimes you don't want to be too precise with your estimates. A loose estimate with a nice form that gets the job done is often easiest.

Here's a start:

$|\frac{n^2+1}{n^2+n+1}-1|=\frac{n}{n^2+n+1}<\frac{n}{n^2}=\frac1n$

Answer 2

$$\left|\frac{n}{n^2+n+1}\right| <\epsilon\\\left|\frac{n}{n^2+n+1}\right| <\left|\frac{n}{n^2+n}\right| <\left|\frac{n}{n(n+1)}\right| <\left|\frac{1}{n+1}\right| <\epsilon\\n+1 >\frac{1}{\epsilon}\\n>\frac{1}{\epsilon}-1\\n\geqslant \left \lfloor \frac{1}{\epsilon}-1 \right \rfloor $$

Answer 3

$|\frac{-n}{n^{2}+n+1}| < \epsilon \Leftrightarrow |\frac{n}{n^{2}+n+1}| < \epsilon $

Now $|\frac{n}{n^{2}+n+1}| < |\frac{1}{n}| $ . So if $n_0=\frac{1}{\epsilon}$ everything is ok.