Give the \(\varepsilon\)-\(\delta\) definition of the one-sided limit \(\lim_{x \to a^+} f(x) = L\).
Using the definition and basic properties of \(\ln(x)\), show that \(\lim_{x \to 0^+} 1/\ln(x) = 0\).
<1>2. Let \(\varepsilon>0\) and set \[ \delta=e^{-1/\varepsilon}. \] Then \(0<\delta<1\).
<1>3. If \(0<x<\delta\), then \[ \ln x<-\frac1\varepsilon<0. \]
<1>4. Therefore, if \(0<x<\delta\), \[ \left|\frac1{\ln x}-0\right|<\varepsilon. \]
<1>5. Hence \[ \boxed{\lim_{x\to0^+}\frac1{\ln x}=0.} \]
Solution
1. The statement\[ \lim_{x\to a^+}f(x)=L \]means: for every \(\varepsilon>0\) there exists \(\delta>0\) such that\[ 0<x-a<\delta\implies |f(x)-L|<\varepsilon. \]
The logarithm is strictly increasing on \((0,\infty)\). Hence \[ 0<x<e^{-1/\varepsilon} \implies \ln x<\ln(e^{-1/\varepsilon})=-\frac1\varepsilon. \]
By 3, \(\ln x<0\) and \(|\ln x|>1/\varepsilon\). Thus \[ \left|\frac1{\ln x}\right| =\frac1{|\ln x|} <\varepsilon. \]
The estimate in 4 holds for every \(x\) satisfying \(0<x-0<\delta\), so it is exactly the right-hand \(\varepsilon\)-\(\delta\) condition from 1.