Find the limit $\lim_{n \to \infty} \frac {2n^2+10n+5}{n^2}$ and prove it.

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 $\frac1n \to 0$.

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:

$$\forall\varepsilon>0,\exists M\in\mathbb{Z}^+,\forall n\ge M,|u_n-l|<\varepsilon.$$