Prove, using only the definition of $O()$, that $2^{\sqrt{x}}$ is not $O(x^{10})$

Question

Prove, using only the definition of $O()$, that $2^{\sqrt{x}}$ is not $O(x^{10})$.

I have been doing a few exercises on Big O and this is the first time I have encountered the variable in the exponent. I was wondering how to disprove this function. Any help would be appreciated

Answer 1

Hint: Try converting both functions to the form $e^{f'(x)}$.

Answer 2

$O(x^{10})$ is $O(x^k)$ for some constant $k$ (polynomial complexity) whereas $O(2^{\sqrt{x}})=O(2^{x^k})$ is subexponential by definition.

Using De l'Hôpital's rule, you can find the following:

$ \lim_{x\rightarrow\infty} \displaystyle\frac{x^{10}}{2^{\sqrt{x}}} = \lim_{x\rightarrow\infty} \frac{\frac{\delta x^{10}}{\delta x}}{\frac{\delta 2^{\sqrt{x}}}{\delta x}}=0 $

which shows that the denominator grows faster than the numerator. You can actually differentiate the fraction until you get a constant numerator while the denominator will always stay a function of $x$.