## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0^+}\frac{\ln x}{x}=-\infty$$
$$A=\lim_{x\to0^+}\frac{\ln x}{x}$$ As $\lim_{x\to0^+}(\ln x)=-\infty$ (as $x\to0^+$, $\ln x\to-\infty$) and $\lim_{x\to0^+}x=0$ this is a form of $\frac{-\infty}{0}$, so L'Hospital's Rule cannot be applied. However, we can still judge intuitively. As $x\to0^+$, $\ln x\to-\infty$, a negatively very large number, while $x\to0^+$, a very small positive number (but never reaches $0$). So we see here that the limit of the numerator has already reached negative infinity. When being divided by a very small positive number, it would become even smaller, going further into negative infinity. In other words, $$A=\lim_{x\to0^+}\frac{\ln x}{x}=-\infty.$$