Apologize if the question sounds absurd, but what is the behavior of $\log x$ around $x=0$? I can see that it dominates $-\frac1x$, but what about for other negative powers of $x$, for instance $-\frac{1}{\sqrt{x}}$?
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2For any $a>0$, $x^a\log(x)\to 0$ as $x\to 0$. – Mark Viola May 26 '18 at 15:04
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Using only the limit definition of the exponential function and Bernoulli's Inequality, I showed in THIS ANSWER that the logarithm function satisfies the inequality
$$\log(x)\ge \frac{x-1}{x}>-\frac1x\tag1$$
Since $a\log(x)=\log(x^a)>-\frac1{x^a}$, for any $a>0$ we have from $(1)$
$$\log(x)>-\frac1{ax^a}$$
Mark Viola
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If $r>0$, then, by L'Hopital's rule,$$\lim_{x\to0^+}\frac{\log x}{-\frac1{x^r}}=\lim_{x\to0^+}\frac{x^{-1}}{rx^{-r-1}}=\frac1r\lim_{x\to0^+}x^r=0.$$
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José Carlos Santos
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