Hi I attach my question.
In this I have assumed Archimeadan's property, there probably are other more efficient ways but I think what I have done is correct, if anyone could take a look at it- it would be awesome thnx
It's better to look at these limits: $$ \lim_{n \to +\infty}{5(1+\frac{1}{n})}-\lim_{n \to +\infty}{2(1+\frac{1}{n^2})}= \lim_{n \to +\infty}{5}+\lim_{n \to +\infty}{\frac{5}{n}}-\lim_{n \to +\infty}{2}+\lim_{n \to +\infty}{\frac{2}{n^2}}=5+0-2 $$
No need for that. You know $\lim_{n\to\infty}\frac{1}n=0$ and $\lim_{n\to\infty}\frac{1}n\cdot \frac{1}n =$$\lim_{n\to\infty}\frac{1}n. \lim_{n\to\infty}\frac{1}n =0\cdot0=0$ Now use limit of sum is sum of limits for convergent sequences and you're done.
$\lim_{n\to\infty}5(1+\frac{1}n)- 2\cdot\lim_{n\to\infty}(1+\frac{1}n\cdot \frac{1}n)= 5-2=3$
This looks correct although the step where you remove the absolute value looks like you could add some more justification.
Some other pointers:
You could just say for N := $\lceil \frac{3}{\epsilon} \rceil$, then n>N implies $\frac{1}{n} < \frac{\epsilon}{3}$ to avoid referring to the property.
You could also pull a $\frac{1}{n}$ out of the original absolute value and have $\frac{1}{n}\left(5-\frac{2}{n}\right) < \frac{5}{n}$ at the end of your second line and just conclude the proof by setting N:=$\lceil\frac{5}{\epsilon}\rceil$