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Using l'hopital rulelim x -> inf ln(x^2)/ln(x^3)

User Troggy
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1 Answer

1 vote

Given data:

The given expression.


\lim _(x\to\infty)(\ln x^2)/(\ln x^3)

The given expression can be written as, after applying the L'Hospital rule.


\begin{gathered} \lim _(x\to\infty)(\ln x^2)/(\ln x^3)=\lim _(x\to\infty)((d)/(dx)(\ln x^2))/((d)/(dx)(\ln x^3)) \\ =\lim _(x\to\infty)(((1)/(x^2))(2x))/(((1)/(x^3))(3x^2)) \\ =\lim _(x\to\infty)(2)/(3) \\ =(2)/(3) \end{gathered}

Thus, the value of limits of the given expression is 2/3.

User Neisha
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