<s>[INST] Proving the limit of a sequence using definitions This is a review that my professor posted and I want to make sure I'm on the right path as I study


*

*$\cdot \lim \limits_{n \to \infty} n - \sqrt{2n^2+1} = n -\sqrt{2n^2+1}*\frac{n+\sqrt{2n^2+1}}{n+\sqrt{2n^2+1}} = \frac{-n^2-1}{n+\sqrt{2n^2+1}}$


So can I say that $M<\frac{-n^2-1}{n+\sqrt{2n^2+1}} \le \frac{n^2}{2n}\le \frac{n}{2}   $ and let my function be $N(M) = \frac{2}{M}$ when $n>N$


*$\cdot \lim \limits_{n \to \infty} \frac{10^n}{n!} $ so can I say for a large n, maybe $n>20$  $\frac{10^n}{n!}\le \frac{1}{n!}$ therefore the function $N(\epsilon) = \max${$20, \epsilon!$} I feel like I'm very off track with this one

*This is the reverse of 2, but wouldn't it be the same approach but to find $N(M) $ such that $M>n!$ 

*Finally,
$\cdot \lim \limits_{n \to \infty} \frac{n}{3n-2000} = \frac{\frac{n}{n}}{\frac{3n}{n}-\frac{2000}{n}} = \frac{1}{3-\frac{2000}{n}}$ so then find $N(M)$ such that $M> \frac{1}{3} - \frac{n}{2000}$ therefore $N(M) = 2000(M-\frac{1}{3})$ or would $N(M) = Max${$\frac{2000}{3}, 2000M$}
 [/INST] For the first sequence: If $n \geq 1$, then
$$
n - \sqrt{2n^{2}+1} = \frac{n^{2}-2n^{2}-1}{n+\sqrt{2n^{2}+1}} = \frac{-n^{2}-1}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n + \sqrt{2n^{2}+1}} < \frac{-n^{2}}{n+4n} = \frac{-n}{5};
$$
given any $M < 0$, we have $\frac{-n}{5} < M$ if $n > 5|M|$; so for all $n \geq \lceil 5|M| \rceil +1$ we have $n - \sqrt{2n^{2}+1} < M$.
For the second sequence: If $n \geq 11$, then
$$
\frac{10^{n}}{n!} = \frac{10^{n}}{1\cdot 2 \cdot \cdots n} \leq \frac{10^{10}}{10!}\cdot \frac{10}{n} = \frac{10^{10}}{9!n};
$$
taking any $\varepsilon > 0$, we have $10^{10}/9!n < \varepsilon$ if $n \geq \lceil \frac{10^{10}}{9!\varepsilon} \rceil + 1$; so if $n \geq \max \{11, \lceil \frac{10^{10}}{9!\varepsilon} \rceil + 1 \}$, then $\frac{10^{n}}{n!} < \varepsilon$.
For the third sequence, by initial inspection we may guess that the sequence converges to $\frac{1}{3}$. To prove this, note that, if $n \geq 1$, then 
$$
\bigg| \frac{n}{3n-2000} - \frac{1}{3} \bigg| = \frac{2000}{9n-6000};
$$
taking any $\varepsilon > 0$, if $n \geq \lceil \frac{2000(1+3\varepsilon)}{9\varepsilon} \rceil + 1$, then $\frac{2000}{9n-6000} < \varepsilon$.
</s>