Find the limit (proof). (assignment) This is my assignment question.
Find $\lim_{n \to \infty} (2n^2+10n+5)/n^2$.  
My attempt: suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$ 
Then, $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}$.  
Then, let $M \in N$ and $\frac {15}M< \varepsilon$. Then, for $\varepsilon >0$, there exists $M$ such that $|\frac {2n^2+10n+5}{n^2}-2|= \frac {10n+5}{n^2}\le \frac {15n}{n^2}= \frac {15}{n}\le\frac {15}{M}<\varepsilon$ for $n\ge M$.
My question is (1) Can I say "suppose $\lim_{n \to \infty} (2n^2+10n+5)/n^2=2$" before finding the limit ?
(2)$\frac {10n+5}{n^2}\le \frac {15n}{n^2}$. I am also not sure whether it is true. 
Thank you in advance.
 A: You could just do $$\lim_{n \to \infty}\frac{2n^2+10n+5}{n^2}=\lim_{n\to\infty}\frac{2+\frac{10}n+\frac5{n^2}}{1}=\frac{2+0+0}{1}=2$$ since both $\dfrac1n$ and $\dfrac1{n^2}$ tend to $0$ as $n\to\infty$.
A: (1) You can say: " I claim that the limit $=2$".
Then you prove this claim. Your proof above is fine.
(2) for $n \in \mathbb N$ we have $5 \le 5n$ , hence $10n+5 \le 15n$ and $\frac {10n+5}{n^2}\le \frac {15n}{n^2}$ follows.
A: Of course you can, because you never use this assumption in the proof !
You are indeed working on $|f(n)-2|$ and come to establish it $ <\epsilon$, but nowhere do you use the fact that $\lim f(n)=2$.
Here $2$ is just a tentative value (which you found by educated guess), and you prove it to be the right choice.

If there was a line in your proof saying "knowing that $\lim f(n)=2$, then ...", this would of course be a circular argument.
A: Yes of course we need to claim that the limit is equal to 2, indeed this kind of proof are not used for the calculation of the limit but to get confident with the definition and to prove the simplest base case
For the proof the logic is firstly fix $\epsilon>0$ and then find $N$ such that for $n\ge N$
$$|\frac {2n^2+10n+5}{n^2}-2|<\epsilon\implies10n+5<\epsilon n^2\implies\epsilon n^2-10n-5>0 \\\implies n>\frac{10+\sqrt{100+20 \epsilon}}{2\epsilon}$$
which complete the proof.
