Help with Epsilon-Delta proof of Limit at infinity Quotient

Question

I am having difficulty with this epsilon-delta proof.
Since my L here is zero, do I solve for delta in the numerator and denominator separately? Any help would be appreciated!

Answer 1

Note that for $n>2$ we have $n^2-2>n^2/2$.

Thus if $n>2$ then $$n/(n^2-2) < \frac {n}{n^2/2} = 2/n $$ Let $\epsilon >0$ be an arbitrary small number. Choose a natural number $n_0$ such that $n_0 > 2/\epsilon .$

If $n\ge \text {max {2,$n_0$}},$ we have $2/n\le 2/n_0 <\epsilon.$ That is, $lim _{x\to \infty} \frac {n}{n^2-2} =0$

Answer 2

Two things to remember:

i) In order to bound a quotient from above (like is required here), you need to bound the denominator from below and the numerator from above.

ii) Given one $\varepsilon>0$ you need to find one $n_0$ that works. But you don't need to find the optimal one (which probably would be much more difficult in any non trivial example). Just one that works!

In this case, the main idea is that for $n$ large enough, $n^2-2$ will behave more or lees line $n^2$ (I mean, it is of the same order of magnitud ), and this is why the bound suggested by another user

$$ n^2-2 > \frac{1}{2} n^2 $$

is good. This hold indeed for $n> 2$ as you can easily check, for instance by plotting the function $f(x)=x^2-x/2-2$, or by computing the roots of this quadratic polynomial.

Then $$ |\frac{n}{n^2-2}| \leq \frac{n}{n^2/2} = \frac{2}{n} $$ which is less than $\varepsilon$ provided that $n> \min\{\frac{2}{\varepsilon},2 \} $