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Why is $(\log(n))^{99} = o(n^{\frac{1}{99}})$
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fracwhylog
Problem
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.
$$
\lim_{n \to \infty} \frac{ (\log(n))^{99} }{n^{\frac{1}{99}}}
$$
But I'm not sure how I can reduce this expression.
Solution
$\qquad \begin{align}
\lim_{x \to \infty} \frac{ (\log(x))^{99} }{x^{\frac{1}{99}}}
&= \lim_{x \to \infty} \frac{ (99^2)(\log(x))^{98} }{x^{\frac{1}{99}}} \\
&= \lim_{x \to \infty} \frac{ (99^3) \times 98(\log(x))^{97} }{x^{\frac{1}{99}}} \\
&\vdots \\
&= \lim_{x \to \infty} \frac{ (99^{99})\times 99! }{x^{\frac{1}{99}}} \\
&= 0
\end{align}$
I used L'Hôpital's rule law in each conversion assuming natural logarithm.
\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.
Context
StackExchange Computer Science Q#9852, answer score: 7
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