Comparing the growth rate of two functions using L'Hopital's rule

Question

For the two functions $f(n)=n^{100}$ and $g(n)=2^{n/100}$, I am trying to determine whether $f(n) = O(g(n))$. In order to do this, I used L'Hopital's rule as if $\lim\limits_{n\to\infty} \frac{f(n)}{g(n)}=0$, this implies that eventually $f(n) < g(n)$, and thus $f(n) = O(g(n))$.

As the following limit is of indeterminate form, L'Hopital's rule can be used: $$\lim_{n\to\infty} \frac{f(n)}{g(n)} = \frac{\infty}{\infty}$$

However, when I tried using L'Hopital's rule using the derivatives of $f(n)$ and $g(n)$ with respect to $n$, the limit is of indeterminate form: $$\lim_{n\to\infty} {100n^{99}\over 2^{n/100-2}\cdot \ln2/25} = \frac{\infty}{\infty}$$

Is there any other method I can use to demonstrate the $f(n) = O(g(n))$, since using L'Hopital's rule did not work. Any insights are appreciated.

Answer 1

Nor mally you shouldn't even think of L'Hospital? It is a basic result that for any $\alpha, \beta>0$, one has $$x^\alpha =o(\mathrm e^{\beta x}) \quad(x\to+\infty),\;\text{ which means }\lim_{x\to+\infty}\frac{x^\alpha}{\mathrm e^{\beta x}}=0.$$ Now, note that $2^{n/100}=\mathrm e^{\ln 2\cdot n/100}$.

Answer 2

We have $\log\left(n^{100}\right)=100\log(n)$ and $\log\left(2^{n/100}\right)=\frac{\log(2)}{100}n$. Both of these tend to $\infty$. Applying L'Hopital to the logs gives $$ \begin{align} \lim_{n\to\infty}\frac{100\log(n)}{\frac{\log(2)}{100}n} &=\lim_{n\to\infty}\frac{100\frac1n}{\frac{\log(2)}{100}}\\ &=0 \end{align} $$