Given data:
The given expression.
[tex]\lim _{x\to\infty}\frac{\ln x^2}{\ln x^3}[/tex]The given expression can be written as, after applying the L'Hospital rule.
[tex]\begin{gathered} \lim _{x\to\infty}\frac{\ln x^2}{\ln x^3}=\lim _{x\to\infty}\frac{\frac{d}{dx}(\ln x^2)}{\frac{d}{dx}(\ln x^3)} \\ =\lim _{x\to\infty}\frac{(\frac{1}{x^2})(2x)}{(\frac{1}{x^3})(3x^2)} \\ =\lim _{x\to\infty}\frac{2}{3} \\ =\frac{2}{3} \end{gathered}[/tex]Thus, the value of limits of the given expression is 2/3.
