Answer
$$\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}=0$$
Work Step by Step
$$A=\lim_{x\to0^+}\frac{\arctan(2x)}{\ln x}$$
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, $$A=0$$
