Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$
 A: The following calculation may be helpful:
$$\left|\frac{n^2-2}{3n^2+1}-\frac13\right| = \left|\frac{(3n^2-6)-(3n^2+1)}{3(3n^2+1)}\right| = \frac73\left(\frac1{3n^2+1}\right)$$
Can you argue that $3n^2+1$ gets large as $n\to\infty$?
A: For $\epsilon >0$ sufficiently small $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon \Leftrightarrow \left|\dfrac{-7}{9n^2+3}\right|<\epsilon $$ $$ \Leftrightarrow \dfrac{7}{9n^2+3}<\epsilon \Leftrightarrow \frac{7}{\epsilon}-3<9n^2\Leftrightarrow \frac{7-3\epsilon}{9\epsilon}<n^2\Leftrightarrow \sqrt{\frac{7-3\epsilon}{9\epsilon}}<n$$ That is, for every $\epsilon >0$ sufficiently small if we choose $$n_0=\left\lfloor\sqrt{\frac{7-3\epsilon}{9\epsilon}}\right\rfloor +1$$ then $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon $$ if $n\ge n_0$ so, $$\lim_{n\to +\infty}\dfrac{n^2-2}{3n^2+1}=\dfrac{1}{3}.$$
A: I'll help get you started. You want to show that N>n => |(n^2 - 2)/(3n^2 + 1) -1/3| < epsilon. (Definition of convergence of a sequence). 
First, get a common denominator. 
|-7/(9n^2+3)|< epsilon
Then you're gonna want to solve for n in terms of epsilon and set N equal to the value you get for n. 
A: Okay, so lets look at the problem:
$$\lim_{n\to \infty} \frac{n^2 -2}{3n^2 +1}.$$
The key here is to do a bit of algebra - on that basis, remove a factor of $n^2$ on the top and bottom:
$$\lim_{n\to \infty} \frac{n^2\big(1-\frac{2}{n^2}\big)}{n^2\big(3 +\frac{1}{n^2}\big)} = \lim_{n\to \infty} \frac{\big(1-\frac{2}{n^2}\big)}{\big(3 +\frac{1}{n^2}\big)}.$$
Now we can apply the limit laws step by step (I don't know how familiar you are with this, so I'll go slowly), the first is that we can pass the limit over the fraction:
$$\lim_{n\to \infty} \frac{\big(1-\frac{2}{n^2}\big)}{\big(3 +\frac{1}{n^2}\big)} = \frac{\lim_{n\to \infty} \big(1-\frac{2}{n^2}\big)}{\lim_{n\to \infty} \big(3 +\frac{1}{n^2}\big)}.$$
Then we can pass the limit through the brackets (baring in mind that the limit of a constant is just the constant):
$$\frac{\big(1-\lim_{n\to \infty} \big[ \frac{2}{n^2}\big]\big)}{ \big(3 +\lim_{n\to \infty}\big[ \frac{1}{n^2}\big]\big)}.$$
Knowing now that $\lim_{n\to \infty} \frac{1}{n^2} = 0$ we have that
$$\lim_{n\to \infty} \frac{n^2 -2}{3n^2 +1} = \frac{1-0}{3+0} = \frac{1}{3}.$$
A: The Archimedean Property can be used to show that
$\tag 1 \displaystyle \lim_{n \to \infty}\frac{1}{n^2} = 0$
It is easy to show that
$\tag 2\frac{n^2 - 2}{3n^2 + 1} \lt \frac{1}{3} \text{ for all } n \ge 0$
The following statements are all equivalent for any $0 \lt \varepsilon \lt \frac{7}{3}$ when $n \gt 0$:
$\tag 3 \frac{1}{3} - \varepsilon \lt \frac{n^2 - 2}{3n^2 + 1}$
$\tag , n^2 + \frac{1}{3} - 3 n^2 \varepsilon  - \varepsilon \lt n^2 - 2$
$\tag , \frac{7}{3}  - \varepsilon \lt 3 n^2 \varepsilon$
$\tag 4 \frac{1}{n^2}  \lt \frac{3 \varepsilon}{\frac{7}{3}  - \varepsilon}$
By (1) we can find an $N$ so that (4) is true when $n \ge N$. But then both (2) and (3) show that
$\tag 5 \lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$
