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How can I prove that $f(n) > g(n)$, given $f(x)=x^{1+\epsilon}$ and $g(x)=\frac{x^2}{\log_2(x)}$ where $0 < \epsilon < 1$?

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    You mean $f(n)$n$ sufficiently big ? – 2017-02-19
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    You have to find the m such that $f(x)>g(x)$ for all $x>m$ – 2017-02-19
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    yes please. please solve it for me because i'm not mathematician – 2017-02-19
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    @TahaAlmasri: Welcome to Math.SE! Could you please say what tools you have, and a little about where the question comes from? For example, do you know that multiplying both sides of an inequality by a positive function preserves the inequality? Do you know calculus? Do you know how to prove that if $\epsilon > 0$, then $\log_{2}(x)/x^{\epsilon} \to 0$ as $x \to \infty$? Thank you. – 2017-02-19
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    Hello @AndrewD.Hwang the question comes from computer algorithms to determine which algorithm is better than the other one . I know that this problem can be solved with L'Hôpital's rule, which means, I must derive this limit to solve it. But i don't know how ? – 2017-02-19
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    One approach is to compute $\lim\limits_{x\to\infty} f(x)/g(x) = \lim\limits_{x\to\infty} \log_{2}(x)/x^{1-\epsilon}$, which as you say can be accomplished with l'Hopital's rule. (In case it helps, $\log_{2}(x) = \ln x/\ln 2$ for all real $x > 0$.) If this limit if less that $1$, then $f(x) < g(x)$ for sufficiently large $x$. – 2017-02-21

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