How to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I want to show that $\frac{n^3}{\sqrt{n^6 + 1}}$ converges to 1. I've tried using $\epsilon-N$ to show that it converges but I can't isolate $n$. Is there a way to bound this so I can apply the squeeze theorem? Or am I not on the right track?
 A: Yes, we can establish inequalities on the denominator. For sufficiently large $n$, we have
$$
n^3<\sqrt{n^6+1}<\sqrt{n^6+2n^3+1}<n^3+1
$$
As a result, we have
$$
{n^3\over n^3+1}<{n^3\over\sqrt{n^6+1}}<1
$$
By squeeze theorem, we obtain
$$
\lim_{n\to\infty}{n^3\over\sqrt{n^6+1}}=1
$$
A: Hint:
$$\frac{n^3}{\sqrt{n^6 + 1}}=\frac{1}{\sqrt{1+\frac{1}{n^6}}}$$
A: If $\{a_n\}$ is a sequence where $a_n\geq 0$, then $a_n \to L$ if and only if $a_n^2\to L^2$. In our case $a_n^2 = \frac{n^6}{n^6 + 1}$ which is much easier to work with if you want to do an $\epsilon$ proof.
A: If you want to use the sqeeze theorem you can use
$\frac {n^3}{(n+1)^3} = \frac {n^3}{\sqrt{(n+1)^6}}=\frac {n^3}{n^6+ 6n^5 + .... + 1} < \frac {n^3}{\sqrt{n^6 + 1}} < \frac {n^3}{\sqrt{n^6}}=\frac {n^3}{n^3} =1$ (but you will have to prove $\frac {n^3}{(n+1)^3} \to 1$).
But  to do an $\epsilon-N$ proof:
$|\frac {n^3}{\sqrt{n^6 + 1}} - 1| < \epsilon\Leftarrow$
$1-\epsilon < \frac {n^3}{\sqrt{n^6+1}} < 1 + \epsilon$
$1 -2\epsilon + \epsilon^2 < \frac {n^6}{n^6 +1} < 1 + 2\epsilon + \epsilon^2$ (assuming $\epsilon < 1$)
$1-3\epsilon < 1-\epsilon=1-2\epsilon +\epsilon < 1-2\epsilon + \epsilon^2$ and $1+2\epsilon + \epsilon^2 < 1+3\epsilon$.  So if we assumme $\epsilon < \frac 13$ then
$|\frac {n^6}{n^6 + 1}-1| < 3\epsilon$
$|\frac {n^6}{n^6 + 1} -\frac {n^6+1}{n^6+1}| < 3\epsilon$
$|-\frac {1}{n^6+1}|< 3\epsilon$
$1 < 3(n^6 +1) \epsilon$
$\frac 1{3\epsilon} -1 < n^6$ so if $N \ge \sqrt[6]{\frac 1{3\epsilon'} -1}$ (where $\epsilon' = \min(\epsilon, \frac 13)$ we are done.
$n > N\implies n^6 > \frac 1{3\epsilon} -1\implies |\frac {n^3}{\sqrt{n^6+1} -1}|<\epsilon$.
