Question

Calculate each of the limits in Exercises $21-48$. Some of these limits are made easier by L’Hopital’s rule, and some are not.

$\underset{x\to \infty }{\lim }\frac{\ln x}{\ln (2x+1)}$.

Short answer

The value of the limit$\underset{x\to \infty }{\lim }\frac{\ln x}{\ln (2x+1)}$is,$0.$

Step-by-step solution

Step 1. Given information

$\underset{x\to \infty }{\lim }\frac{\ln x}{\ln (2x+1)}$.

Step 2. The limit using L'Hopital's rule is given below:

$\underset{x\to \infty }{\lim }\frac{\ln x}{\ln (2x+1)}=\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{\frac{1}{2x+1}}$[L'Hopital's rule]

$$\begin{gathered} =\underset{x\to \infty }{\lim }\left(\frac{1}{x}+\frac{1}{2x+1}\right) \\ =\underset{x\to \infty }{\lim }\left(\frac{1}{2x^2+x}\right) \\ =\frac{1}{\infty } \\ =0 \end{gathered}$$

The value of the limit is,$0$.