Answer
\[\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}=0\]
Work Step by Step
Let \[l=\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}\]
Which is $\frac{0}{0}$ form
Using L'Hopital's rule
\[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{(\tan^{-1} 2x)'}{(\ln x)'}\]
\[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{\frac{2}{1+4x^2}}{\frac{1}{x}}\]
\[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{2x}{1+4x^2}\]
\[\Rightarrow l=\lim_{x\rightarrow 0^+}\frac{2x}{1+4x^2}=0\]
Hence, \[\lim_{x\rightarrow 0^+}\frac{\tan^{-1} 2x}{\ln x}=0\]