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