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

Problem Detail: 

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.

Asked By : David Faux

Answered By : Reza

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

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

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Question Source : http://cs.stackexchange.com/questions/9852

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