Chapter 9.2, Problem 66E

To determine

To find : The value of $\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{3}x}$

Answer to Problem 66E

  $\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{3}x}=\frac{\ln 3}{\ln 2}$

Answer (B).

Explanation of Solution

Given information :

  $\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{3}x}$

Formula used : L’Hospital rule.

Calculation :

  $\begin{array}{l}{\log }_{2}x=\frac{{\log }_{e}x}{{\log }_{e}2}=\frac{\ln x}{\ln 2}\\ {\log }_{3}x=\frac{{\log }_{e}x}{{\log }_{e}3}=\frac{\ln x}{\ln 3}\\ Now,\\ \underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{3}x}=\underset{x\to \infty }{\lim }\frac{\ln x\times \ln 3}{\ln 2\times \ln x}=\frac{\ln 3}{\ln 2}\end{array}$

Therefore, $\underset{x\to \infty }{\lim }\frac{{\log }_{2}x}{{\log }_{3}x}=\frac{\ln 3}{\ln 2}$