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.
 A: In (1),
$$
X_n = \frac{\sqrt{n^2 + 1}}{\sqrt n} = \sqrt{n + \frac 1n} > \sqrt n
$$
for $n > 0$. Also, $X_{n+1} > X_n$, as can be seen from
$$
\begin{align}
\frac{X_{n+1}}{X_n} &= \sqrt{\frac{n+1+\frac{1}{n+1}}{n+\frac 1n}} \\
&= \sqrt{\left(\frac{(n+1)^2 + 1}{n^2 + 1}\right)\frac{n}{n + 1}} \\
&= \sqrt{\frac{n^3 + 2n^2 + 2n}{n^3 + n^2 + n + 1}} \\
&= \sqrt{1 + \frac{n^2 + n - 1}{n^3 + n^2 + n + 1}} \\
&= \sqrt{1 + \frac{(n + \frac{1 + \sqrt{5}}{2})(n + \frac{1 - \sqrt{5}}{2})}{n^3 + n^2 + n + 1}} \\ &> 1
\end{align}
$$
for $n > \frac{\sqrt 5 - 1}2$ (which obviously holds for $n \ge 1$).
Therefore, for any given $\alpha > 0$, the choice $N = \lceil \alpha \rceil^2$ makes $X_n > X_N > \sqrt N = \lceil\alpha\rceil \ge \alpha$ for all $n > N$.
For (2),
$$
X_n = -n + \sin n \le -n + 1
$$
for all $n > 0$. For any given $\beta < 0$, pick $N = -\lfloor \beta \rfloor + 1$. Then, for any $n > N$, we have
$$
X_n \le -n + 1 < -N + 1 = \lfloor \beta \rfloor \le \beta.
$$
A: 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.
