Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}=0$$
$$A=\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}$$ As $x$ approaches $0^+$, $\arctan(2x)$ approaches $\arctan(2\times 0)=\arctan0=0$ However, at the same time, $\ln x$ approaches $\ln0^+$, meaning it approaches $-\infty$. So there is no indeterminate form here, and L'Hospital's Rule cannot be applied. Nevertheless, we see that while the denominator approaches a madly big number, the numerator instead approaches $0$. That means the whole formula still reaches $0$ in the end, for $\frac{0}{a}=0$ ($a$ stands for any real number) In other words, $$A=0$$