Proving that $f(x)=2^x$ is $O(x^2)$ Can someone help me with this problem? I don't really know what to do if the x is in exponential form. 

Suppose that there is $M\in\mathbb{R}$ such that $2^x\leqslant M\cdot x^2,\;\forall\; x\geqslant x_0$. 
That is suppose that $2^x\in O(x^2)$.
Then, $\dfrac{e^{x\log2}}{x^2}\leqslant M,\;\forall\; x\geqslant x_0$. Let $x\rightarrow \infty$, then $\lim_{x\to\infty}\dfrac{e^{x\log2}}{x^2}=\infty\leqslant M$. Absurd.
Hence, $2^x\not\in O(x^2)$.
