Is this computation erroneous?

Question

The solution to $\lim_{n\to\infty}\frac{n^{100}}{2^{n/100}}$ should be $\infty$ because $n^{100}$ grows much faster than $2^{n/100}$

But if you input

lim(n^100/(2^(n/100)),n,inf)

into Mathematica, you get that the limit is 0 as n approaches infinity. Any way to reconcile these results?

Answer 1

It's true that $n^{100}$ is bigger than $2^{n/100}$ when $n=1$ or $100$, or even $1000$, but try $n=1{,}000{,}000=10^6$. You get $n^{100}=10^{600}$ while

$$2^{n/100}=2^{10{,}000}=(2^{10})^{1000}\gt(10^3)^{1000}=10^{3000}$$

Similarly, $n=1{,}000{,}000{,}000=10^9$ gives $n^{100}=10^{900}$ versus $2^{n/100}\gt10^{3{,}000{,}000}$.

Convinced yet?

Answer 2

Use L'hopital rule $100$ times, it's evident that $n$ in numerator vanishes while denominator will still be $2^{n/100} \times $ some factor.