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 $N(M) = Max${$\frac{2000}{3}, 2000M$}
 A: 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 $\frac{2000}{9n-6000} < \varepsilon$.
A: *

*You are good up the end of the line. (Except for the first equality where you have the limit on the left, but the bare function on the right. These are not equal.) But what you've written below is nonsense. What is M, and how can it possibly be less than the fraction for all sufficiently large $n$? You cannot not just assume that some such M exists. In this case, it does not. Instead, I suggest dividing the top and bottom of your fraction by $n$. A little work will make it clear that this limit is $-\infty$ as André Nicolas has said.

*Please look at this supposed inequality you wrote down. The only way for it to be true is if $10^n < 1$. $n > 20$ certainly isn't going to make that true!

*What do you mean by "the reverse of 2"?

*$\frac {1}{3 - \frac{2000}n} \ne {1 \over 3} - \frac{n}{2000}$
