Show $\frac{x^{3} + 2x}{2x + 1}$ is $O(x^2)$
Can I do it like this?
Since exponent rules/laws allow this:
$\frac{x^{3} + 2x}{2x + 1}$ $=$ $\frac{1}{2}x^{2} + 2x$
Must show a constant c>0 and k s.t
$\frac{1}{2}x^{2} + 2x$ $< c * x^{2}$
Consider $\frac{1}{2}x^{2} + 2x^{2}$ $=$ $\frac{1}{2}x^{2} + \frac{4}{2}x^{2}$
$=$ $x^{2}$($\frac{1}{2}$ $+$ $\frac{4}{2}$) = $x^{2}$$(\frac{5}{2})$
So choose c = $\frac{5}{2}$ => $\frac{x^{3} + 2x}{2x + 1}$ < $\frac{5}{2}$$x^{2}$ for any $x>k$ where $k=0$
$\frac{x^3+2x}{2x+1}\le\frac{x^3+2x}{2x}=\frac{1}{2}x^2+1\le 2x^2$ for all $x\ge 1$