Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it. Find $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$.
I claim that $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$. To prove this, for given $\varepsilon >0$, I have to find $M\in N$ such that $|\frac {2n^2+10n+5}{n^2}-2|<\varepsilon$ for $n \ge M$.
By Archimedean property, we can find $M \in N$ such that $\frac {15}M<\varepsilon$, and note that $n\ge M \rightarrow \frac 1n \le \frac 1M \rightarrow \frac {15}n \le \frac {15}M$.
Then, for $n \ge M$, we have that $|\frac {2n^2+10n+5}{n^2}-2|=|\frac {10n+5}{n^2}| < |\frac {15n}{n^2}|$ (since $n \ge M \in N$) $<\frac {15}n\le \frac {15}M<\varepsilon$.
Therefore, by definition of convergence, $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}=2$.
Should I say $M\in Z^+$ (because I am worried about the case where $M=0$)? Can you find any mistakes in this proof?
Thank you in advance.
 A: Your proof is correct, well done.
Some authors don't include $0$ in $\mathbb{N}$, but based on your question seems $\mathbb{N}=\{0,1,2,\dots\}$ for you.
I guess the definition of convergence of a sequence $(u_n)_{n\in\mathbb{N}}$ to $l\in\mathbb{R}$ that you have is this:
$$\forall\varepsilon>0,\exists M\in\mathbb{N},\forall n\ge M,|u_n-l|<\varepsilon.$$
Let $\varepsilon>0$ and $M\in\mathbb{N}$ (that exists) such that $$\forall n\ge M,|u_n-l|<\varepsilon.$$
Notice that, in particular, $$\forall n\ge M+1,|u_n-l|<\varepsilon$$ because $M+1>M$. So you can replace $M$ by $M+1$ or any larger integer. Therefore you can always choose $M$ to be positive. So you can safely have this as a definition:
$$\forall\varepsilon>0,\exists M\in\mathbb{Z}^+,\forall n\ge M,|u_n-l|<\varepsilon.$$
A: Your proof is fine, to simplify note that
$$\frac {2n^2+10n+5}{n^2}=2+\frac{10}{n}+\frac5{n^2}$$
then it suffices to prove that $\frac1n \to 0$.
A: Another possibility: apply two times Cesàro-Stolz.
$$
\lim_{n\to\infty}\frac{2n^2 + 10n + 5}{n^2} =
\lim_{n\to\infty}\frac{((2(n+1)^2 + 10(n+1) + 5) - (2n^2 + 10n + 5)}{(n+1)^2 - n^2} =
\lim_{n\to\infty}\frac{4n + 12}{2n + 1} =
\lim_{n\to\infty}\frac{(4(n+1) + 12) -(4n + 12)}{(2(n+1) + 1) - (2n + 1)} =
\lim_{n\to\infty}\frac42 = 2.
$$
