Why is $(\log(n))^{99} = o(n^{\frac{1}{99}})$

Question

I am trying to find out why $(\log(n))^{99} = o(n^{\frac{1}{99}})$. I tried to find the limit as this fraction goes to zero.

$$ \lim_{n \to \infty} \frac{ (\log(n))^{99} }{n^{\frac{1}{99}}} $$

But I'm not sure how I can reduce this expression.

Answer 1

$\qquad \begin{align} \lim_{x \to \infty} \frac{ (\log(x))^{99} }{x^{\frac{1}{99}}} &= \lim_{x \to \infty} \frac{ (99^2)(\log(x))^{98} }{x^{\frac{1}{99}}} \\ &= \lim_{x \to \infty} \frac{ (99^3) \times 98(\log(x))^{97} }{x^{\frac{1}{99}}} \\ &\vdots \\ &= \lim_{x \to \infty} \frac{ (99^{99})\times 99! }{x^{\frac{1}{99}}} \\ &= 0 \end{align}$

I used L'Hôpital's rule law in each conversion assuming natural logarithm.

Answer 2

Hint: take $99$th root of both sides.

Answer 3

Try a common trick: express both functions as $e^{\dots}$ and compare the exponents; if their ratio tends to $0$ or $\infty$, so does the original quotient (since $e^x$ is monotone).

$\frac{(\log n)^{99}}{n^{\frac{1}{99}}} = \frac{e^{99 \cdot \log(\log n)}}{e^{\frac{1}{99} \cdot \log n}}$