<s>[INST] Show whether $ x^3$ is $O(g(x))$ where $g(x)=x^2 $ I know that $x^3$ is $O(x^3)$ or any exponent higher than 3, but how do I show mathematically that $x^3$ is not $O(x^2)$?
 [/INST] If you want to show that it's NOT the case that $x^3=O(x^2)$, then you are showing that $x^2=o(x^3)$, or in other words, that $x^2$ is "little-oh" of $x^3$, which communicates that $x^2$ is of a smaller order of growth than $x^3$.
As gimusi has noted, you could show $x^2=o(x^3)$ via limits as follows
$$\lim_{x\rightarrow \infty} \frac{x^3}{x^2}=\lim_{x\rightarrow \infty} \frac{3x^2}{2x}=\lim_{x\rightarrow \infty} \frac{6x}{2}=\lim_{x\rightarrow \infty}\frac{6}{2}x=\lim_{x\rightarrow \infty}3x=3\lim_{x\rightarrow \infty}x=3\cdot \infty =\infty$$
*Note the use of of L'Hopital's Rule since $\lim_{x\rightarrow \infty} \frac{x^3}{x^2}=\frac{\infty}{\infty}$ in the beginning

We could also show this using the definition of "big-oh." Consider that $f(x)=O(g(x))$ iff there exists constants $c,k$ such that $|f(x)| \le c \cdot |g(x)|$ for all $x>k$. 
Let $f(x)=x^3$ and $g(x)=x^2$ and assume $x^3=O(x^2)$. Hence, by definition of big-oh, there exists constants $c,k$ such that 
$$|x^3|\le c \cdot |x^2|$$
for all $x>k$. Note that $x^2>0$ and $x^3>0$ as $x\rightarrow \infty$, so we can remove the absolute value signs, having
$$x^3 \le c \cdot x^2$$
Now,
$$\frac{1}{x^2}x^3\le c \cdot x^2\frac{1}{x^2}$$
$$\Rightarrow x \le c$$
By the trichotomy law, for constants $c,k$ we know either $c=k$, $c<k$, or $k<c$.
Case i. $c=k$
By substitution we have $x \le k$. But we already stated $x>k$. Hence, we have contradiction.
Case ii. $c<k$
If $c<k$ and $x \le c$, we have $x < k$. But we already stated $x>k$. Hence, we have contradiction.
Case iii. $k<c$
Recall that $x \le c$ for ALL $x>k$. So whenever $x>c>k$, we have a contradiction.
Thus, if we assume $x^3=O(x^2)$, then in all cases we have a contradiction. Therefore, it is NOT the case that $x^3=O(x^2)$, and it is true that $x^2=o(x^3)$.
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