How to prove this sequence is unbounded? Let $a_{n} = \dfrac{7n^{3} - 3n^{4} -1}{4n^{2} + 3}$
How to show this sequence is unbounded without using limits?
Well I know that I need to show that it unbounded from bottom or above.
I choose bottom, so I need to show that $\forall M \exists n \Rightarrow a_{n}>M$
What is the method? Can someone show me? Thanks!
 A: I have an idea which do not uses limits but i do not know if it helps you.
If $(x_{n})$ is a bounded sequence far from $0,$ that is, if there exists $%
m>0$ such that 
$$
0<m\leq \left\vert x_{n}\right\vert ,\ for\ all\ n\geq n_{0}
$$
for some integer $n_{0},$ and $(y_{n})$ is an unbounded sequence, then the
product $(x_{n}y_{n})$ is an unbounded sequence. 
In fact, assume the contrary, there exist $M>0$ such that 
$$
\left\vert x_{n}\right\vert \left\vert y_{n}\right\vert =\left\vert
x_{n}y_{n}\right\vert \leq M,\ n\geq n_{0}
$$
then%
$$
m\left\vert y_{n}\right\vert \leq \left\vert x_{n}\right\vert \left\vert
y_{n}\right\vert \leq M,\ for\ all\ n\geq n_{0}
$$
therefore%
$$
\left\vert y_{n}\right\vert \leq \frac{M}{m},\ \ \ for\ \ all\ n\geq n_{0}.
$$
It follows that $(y_{n})$ is bounded which is a contradiction with the
hypothesis '' $(y_{n})$ is an unbounded
sequence '' .
To use this idea, it suffices to write your sequence 
$$
\frac{-3n^{4}+7n^{3}-1}{4n^{2}+3}=\frac{-3n^{4}\left[ 1-\frac{7}{3n}+\frac{1%
}{3n^{4}}\right] }{4n^{2}\left[ 1+\frac{3}{4n^{2}}\right] }=\left( \frac{1-%
\frac{7}{3n}+\frac{1}{3n^{4}}}{1+\frac{3}{4n^{2}}}\right) \left( \frac{%
-3n^{2}}{4}\right) =x_{n}y_{n}
$$
where $y_{n}=\left( \frac{-3n^{2}}{4}\right) $ is clearly an unbounded
sequence (in fact, if $M$ is given, then take $n=\left\lfloor \sqrt{\frac{4M%
}{3}}\right\rfloor +1$). Let us proof the existence of $m>0$ such that  $%
x_{n}=\frac{1-\frac{7}{3n}+\frac{1}{3n^{4}}}{1+\frac{3}{4n^{2}}}>m,$ for all 
$n\geq 3.$ In fact 
$1+\frac{3}{4n^{2}}\leq 1+1=2,$ for all $n\geq 1.$ On the other hand, $1-%
\frac{7}{3n}+\frac{1}{3n^{4}}\geq 1-\frac{7}{3n}$ for all $n\geq 1$ and $1-%
\frac{7}{3n}\geq 1-\frac{7}{9}=\frac{2}{9},$ for all $n\geq 3.$ It follows
that%
$$
\frac{1-\frac{7}{3n}+\frac{1}{3n^{4}}}{1+\frac{3}{4n^{2}}}\geq \frac{1-\frac{%
7}{3n}}{1+1}>\frac{1-\frac{7}{9}}{2}=\frac{1}{9}=m>0,\ \ \ for\ \ all\ \
n\geq 3.
$$
A: $$ \frac{7n^{3} - 3n^{4} -1}{4n^{2} + 3}< \frac{7n^{3} - 3n^{4}}{4n^{2} + 3}<\frac{7n^{3} - 3n^{4}}{4n^{2} + 3n^2}=n-\frac{3}{7}n^2= n\left(1-\frac37n\right) \underbrace{\le}_{n\ge 3} 1-\frac37n.$$
Thus, given $M<0$ there exists $N\in\mathbb{N},$ $N\ge 7/3(1-M),$ such that $$n\ge N\implies -n\le -N\le -\frac73(1-M)\implies 1-\frac37 n\le M, $$ from where it follows that $a_n$ is not bounded from below.
It is bounded from above. Indeed, $7n^3-3n^4-1<n^3(7-3n)<0,$ for any $n\ge 3.$ Thus, $$\frac{7n^{3} - 3n^{4} -1}{4n^{2} + 3}<0,$$ for any $n\ge 3.$ 
A: First, note the denominator is always positive, so
$$\frac{7n^3-3n^4-1}{4n^2+3}<0\iff 7n^3-3n^4-1<0\iff -3n^3\left(n-7\right)<1\iff$$
$$3n^3(n-7)>-1$$
and the last inequality is clearly true for $\;n\ge7\;$, and thus the sequence is bounded above, but
$$\frac{7n^3-3n^4-1}{4n^2+3}\le\frac{-3n^4+7n^3}{n^2}=-3n^2+7n$$
which, I believe, is much easier to prove it is unbounded below.
