Don't make mistakes! #5

Sam's teacher gave him a question. The question was:

$\quad$ Find the value of $\displaystyle \lim_{n\rightarrow\infty}\left(\sqrt{n^2-6n+3}-\sqrt{n^2-4n+7}\right)$ .

Now he is going to try and find the answer.

Sam's solution:

$\quad$ 1: $\displaystyle \lim_{n\rightarrow\infty}\frac{2n^2-10n+9}{n^2} = \lim_{n\rightarrow\infty} \frac{2\sqrt{(n^2-6n+3)(n^2-4n+7)}}{n^2}$

$\quad$ 2: $\displaystyle \lim_{n\rightarrow\infty}\left(2n^2-10n+9\right) = \lim_{n\rightarrow\infty} 2\sqrt{(n^2-6n+3)(n^2-4n+7)}$

$\quad$ 4: $\displaystyle \lim_{n\rightarrow\infty}\left(2n^2-10n+10-2\sqrt{(n^2-6n+3)(n^2-4n+7)}\right)=1$

$\quad$ 8: $\displaystyle \lim_{n\rightarrow\infty}\left(\sqrt{n^2-6n+3}-\sqrt{n^2-4n+7}\right)^2=1$

$\quad$ 16: If $\displaystyle n>6,\ \sqrt{n^2-6n+3}<\sqrt{n^2-4n+7}$

$\quad$ 32: Therefore $\displaystyle \lim_{n\rightarrow\infty}\left(\sqrt{n^2-6n+3}-\sqrt{n^2-4n+7}\right)=-1$

Find the sum of all the numbers of the individual equations that are not true. If you think all of the given equations are true, submit 0 as your answer.

---

This problem is a part of <Don't make mistakes!> series .