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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