Showing $\lim_{n \to \infty} \frac{1}{3n^2-2}=0$ and $\lim_{n \to \infty} \frac{1}{6n^2-8n+1}=0$ by the definition of a limit [closed]

Here is part a) worked out, and hopefully you can see how to generalize this approach.

To reiterate, the goal is to show

$$\lim_{n\to\infty} \frac{1}{3n^2 - 2} = 0$$

by the $\varepsilon$-$N$ definition of convergence. That is, for any $\varepsilon > 0$, how to find $N$ (depending on $\varepsilon$) with the property that

$$ \Big| \frac{1}{3n^2 - 2} - 0\Big| < \varepsilon$$

for any $n \geq N$. So to begin attacking this, you might try to start with the above inequality and work backwards. For instance, $3n^2 - 2 > 0$ for every $n = 1, 2, 3, \ldots$, so you can remove the absolute values bars and move terms around to conclude that what you need is $3n^2 - 2 > \varepsilon^{-1}$. As you observed, this is true whenever

$$n > \sqrt{\frac13(2 + \varepsilon^{-1})}$$

So just choose $N$ to be bigger than that number (depending on $\varepsilon$), and you're golden. But this example fools you into doing more than you need to do, and when you do a more general problem, you may find that it is harder to pull off.

See, what this work leads you to believe is this: if you want $f(n) > \varepsilon^{-1}$, then choose $n > f^{-1}(\varepsilon^{-1})$ (note that the first ${}^{-1}$ denotes the inverse of $f$, whereas the second denotes $1/\varepsilon$). But this means you have to go out of your way and compute $f^{-1}$, and then hope that $f$ is such a function that this argument proceeds anyway (in particular, $f$ needs to be strictly increasing and diverge to infinity as $n\to\infty$ in order to work).

There is an easier way, and the easier way boils down to asking less of yourself. It is true that if $n > \sqrt{(2+\varepsilon^{-1})/3}$ then $3n^2 - 2 > \varepsilon^{-1}$. But this bound is the tightest possible, and you don't need such a strong result just to show the limit.

Let's return to this inequality: $3n^2 - 2 > \varepsilon^{-1}$. Now, $\varepsilon$ is the necessary threshold, the fixed constant, the thing whose value you aren't allowed to assume or change. But $n$, on the other hand, is the $free$ constant - you're allowed to choose its value willy-nilly, so long as the above inequality is met.

You're asked to make sure that the quantity $3n^2 - 2$ is very large. But what if you looked at a different quantity? For example, the inequality

\begin{equation}\tag{1} 3n^2 -2 > 3n^2 - n \end{equation}

is true so long as $n > 2$ (i.e., it's true 'for all sufficiently large $n$', the only regime that matters when taking limits). So you could ask, for which $n$ does it hold that $3n^2 - n > \varepsilon^{-1}$? It is a weaker question because if this inequality holds, then by (1) you already have the inequality that you really want (that is, $3n^2 - 2 > \varepsilon^{-1}$).

And that's not the only way you could weaken your question. You could also note that

$$3n^2 - 2 > 3n^2 - n^2 = 2n^2$$

holds for all natural $n > 1$! By dropping some of the lower order terms, it make it easier for me to find how to choose $N$ as a function of $\varepsilon$. With this, I can say something like "choose $n > \sqrt{\varepsilon^{-1}/2}$", and my desired inequality will still be true (so long as $n > 1$).

And you can do even better, still! Note that

$$3n^2 - 2 > n$$

holds for all sufficiently large $n$. I don't know for which specific value it is true off the top of my head (maybe it is true for all $n$), but I know it must be true because the LHS is a polynomial of degree 2, which will grow faster than the RHS (a polynomial of degree 1). For this reason, I can simply say "choose $N > \varepsilon^{-1}$", and then the desired inequality will follow. To show that it really works amounts to simple algebra.

This is the general theme which you should take away from limit proofs. It is not always about finding the tightest or strongest possible bound, but about finding the bound that works.