Real Analysis Question: How to use epsilon-N definition to prove the limit exists.

Question

Here is my question:

Assume $(a_n) \rightarrow 2$. Use $\epsilon-N$ definition to directly prove $lim\frac{1}{a_n^2 - 2} = \frac{1}{2}$.

Here is what I have tried, but I have not gotten very far:

$|\frac{1}{a_n^2}-\frac{1}{2}| = |\frac{2 - a_n^2 + 2}{2(a_n^2-2)}| = |\frac{-a_n^2+4}{2(a_n^2-2)}|$.

I am trying to get the values of $a_n$ into the form $a_n-2$, but I run into trouble with $a_n^2$.

Any advice would be awesome!

Answer 1

Take $\varepsilon>0$. If $n\in\Bbb N$, then$$\frac1{a_n^{\,2}-2}-\frac12=\frac{4-a_n^{\,2}}{2(a_n^{\,2}-2)}=-(a_n-2)\frac{a_n+2}{2(a_n^{\,2}-2)}.\tag1$$Now, take $N_1\in\Bbb N$ such that $n\geqslant N_1\implies|a_n-2|<\frac12$. But then $\frac32<a_n<\frac52$ and therefore $\frac12<2(a_n^{\,2}-2)<\frac{17}2$. On the other hand, $a_n+2<\frac92$. So$$\frac{a_n+2}{2(a_n^{\,2}-2)}<\frac{9/2}{1/2}=9.\tag2$$Now, take $N\in\Bbb N$ such that $N\geqslant N_1$ and that $n\geqslant N\implies|a_n-2|<\frac\varepsilon9$ and it will follow from $(1)$ and $(2)$ that$$n\geqslant N\implies\left|\frac1{a_n^{\,2}-2}-\frac12\right|<\varepsilon.$$

Answer 2

We have that eventually $|a_n-2|\le \frac 12 \iff \frac32\le a_n \le \frac52$ therefore

$$\left|\frac{1}{a_n^2-2}-\frac{1}{2}\right|= \left|\frac{4-a_n^2}{2(a_n^2-2)}\right|=\left|\frac{2+a_n}{2(a_n^2-2)}\right||a_n-2|\le \frac 5 {2\frac14}\cdot |a_n-2|=10|a_n-2|$$

and it suffices to take $n$ such that eventually $|a_n-2|\le \frac {\varepsilon}{10}$.