Question about limit at infinity I have questions which I need to solve:

1) $\lim\limits_{n\to\infty}\dfrac{\sqrt{n^2+1}}{\sqrt n}=\infty$
2) $\lim\limits_{n\to\infty}(\sin(n)-n)=-\infty$

Using this definition:

$X_n\rightarrow\infty \iff$ for all $\alpha>0$, these exists an $N$ in $\Bbb N$ such that if $n>N$, then $X_n > \alpha$
$X_n\rightarrow-\infty \iff$ for all $\beta<0$, these exists an $N$ in $\Bbb N$ such that if $n>N$, then $X_n < \beta$

Please solve it by using these definitions. What are the values for $N$?
Thank you.

I think these are very simple, even if one knows only a little calculus.
For (I), for any $\alpha$, take $n$ greater than the maximal root of $n^2-\alpha^2n+1=0.$ Then, for $m\ge n$, $\frac{\sqrt{m^2+1}}{\sqrt m}\ge \alpha.$
For (II), for every $\beta$, take $n\ge |\beta|+2$, then, for $m\ge n$, $-n+sin(n)\le\beta.$
So the results follow.  
Inform me of any appearance of mistakes, thanks in advance.
