My homework assignment is to proof that $\frac{1}{x\sqrt{x}}$ is $O(\frac{1}{x})$.
I've seen different definitions of Big-Oh, but in my book it is defined as $|f(x)| \leq M \cdot|g(x)|$ for all $x \geq N$ (usually I don't see Big-Oh defined having absolute values for $f(x)$ and $g(x)$).
Now, I could come up with a table of values, but of course that doesn't proof anything. Could I proof it by doing this (I saw it somewhere searching for anwers)? $$ \lim_{x\to\infty}\frac{\frac{1}{x\sqrt{x}}}{ \frac{1}{x}} = \lim_{x\to\infty}\frac{1}{\sqrt{x}}=\sqrt{\lim_{x\to\infty}\frac{1}{x}}=\sqrt{0}=0. $$
Because the limit exists, have I proven that $\frac{1}{x\sqrt{x}}$ is $O(\frac{1}{x})$? How come this works? It looks a whole lot like the limit comparison test. Are there any other ideas which could help me figure out if $\frac{1}{x\sqrt{x}}$ is $O(\frac{1}{x})$?
You proved that there exists $N>0$ such that $$\frac{1}{x\sqrt{x}} \leq \frac{1}{x}$$ for every $x>N$. Just take $\epsilon=1$ in the definitin of limit. By the way, you proved much more, namely that $$\frac{1}{x\sqrt{x}} = o \left( \frac{1}{x} \right)$$ as $x \to +\infty$.