Q. 65 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to \infty }{lim}\left(\ln x^2-\ln (2x+1)\right)$

Step-by-Step Solution

Answer

The limit is infinity

Step 1. Given information

The given expression is  $\underset{x\to \infty }{lim}\left(\ln x^2-\ln (2x+1)\right)$

Step 2. Calculation

The limit is calculated as below, 

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