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.

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 $n\to\infty$.
