1
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

2
$\begingroup$

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 $$ X_n \le -n + 1 < -N + 1 = \lfloor \beta \rfloor \le \beta. $$

$\endgroup$
1
  • $\begingroup$ You're right. I forgot to change that when I put in $\lceil \cdot \rceil$. Thanks. Will edit it now. $\endgroup$
    – Tunococ
    Apr 4, 2013 at 7:27
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ thank you so much for Tunococ and awllower $\endgroup$
    – leena adam
    Apr 4, 2013 at 7:30
  • $\begingroup$ My peasure to help. $\endgroup$
    – awllower
    Apr 4, 2013 at 7:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .