Q. 10 - Calculus

Question

Consider the limit expression

$\underset{x\to \infty }{\lim }\ln \left(\frac{x-1}{1-3x^2}\right)$

Calculate the limit.

Step-by-Step Solution

Answer

The limit  of the expression is $\infty$.

Step 1. Given

The given expression is 

$\underset{x\to \infty }{\lim }\ln \left(\frac{x-1}{1-3x^2}\right)$

Step 2. Calculation

The evaluated expression is as follow:

$$\begin{gathered} \left|\underset{x\to \infty }{\lim }\ln \left(\frac{x-1}{1-3x^2}\right)\right| \\ =\underset{x\to \infty }{\lim }\ln \left(\frac{x^2\left(\frac{1}{x}-\frac{1}{x^2}\right)}{x^2\left(\frac{1}{x^2}-3\right)}\right) \\ =\underset{x\to \infty }{\lim }\ln \left(\frac{\frac{1}{x}-\frac{1}{x^2}}{\frac{1}{x^2}-3}\right) \\ =\ln \left(\frac{\frac{1}{\infty }-\frac{1}{{\infty }^2}}{\frac{1}{{\infty }^2}-3}\right) \\ =\ln \left(\frac{0-0}{0-3}\right) \\ =\ln \left(0\right) \\ =\infty \end{gathered}$$