$\log x$ behaviour near $0$

Question

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 $-\frac{1}{\sqrt{x}}$?

Answer 1

If $r>0$, then, by L'Hopital's rule,$$\lim_{x\to0^+}\frac{\log x}{-\frac1{x^r}}=\lim_{x\to0^+}\frac{x^{-1}}{rx^{-r-1}}=\frac1r\lim_{x\to0^+}x^r=0.$$

Answer 2

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 $(1)$

$$\log(x)>-\frac1{ax^a}$$