Show that $\frac{x^4 +7x^3+5}{4x+1}$ is big-theta($x^3$) I'm having trouble grasping how to set these types of problems.  There are a lot of related questions but it's difficult to abstract a general procedure on finding constants that give the given function bounding constraints to make it big-theta(general function).  
so $\frac{x^4 +7x^3+5}{4x+1}$ is $  \Theta (x^3) $
to show this, we need to find constants such that.
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1} \leq |c_2|(x^3)$$
In addition, there also has to be a $k$ such that for all values $x >k $ the argument holds.
start with one inequality
$$ |c_1|(x^3) \leq \frac{x^4 +7x^3+5}{4x+1}$$
$$ = |c_1| \leq \frac{x^4 +7x^3+5}{4x^4+x^3}$$
$$ = |c_1| \leq \frac{x^4}{x^3(4x+1)} + \frac{7x^3}{x^3(4x+1)} + \frac{5}{x^3(4x+1)}$$
so basically for $x > 0$, $$ |c_1| \leq \frac{1}{4} + 0 + 0$$
I'm assuming after I take the limit as x goes to infinity, i could choose any $c_1$ less than or equal to $\frac{1}{4}$?  The other way would then have the same procedure?  What would I set $k$ to?

Here is a nice simple method.
If $x>1$ then
$$\frac{x^4 +7x^3+5}{4x+1}<\frac{x^4+7x^4+5x^4}{4x}=\frac{13}{4}x^3$$
and
$$\frac{x^4 +7x^3+5}{4x+1}>\frac{x^4}{4x+x}=\frac{1}{5}x^3\ .$$
That is, we have shown that if $x>1$ then
$$\frac{1}{5}x^3<f(x)<\frac{13}{4}x^3\ .$$
