Chapter 9.2, Problem 65E

To determine

To find : $\underset{x\to 1}{\lim }\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}$

Answer to Problem 65E

  $\underset{x\to 1}{\lim }\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}=\frac{1}{2}$

Answer (D).

Explanation of Solution

Given information :

  $\underset{x\to 1}{\lim }\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}$

Formula used : L’Hospital Rule.

Calculation :

  $\begin{array}{l}\underset{x\to 1}{\lim }\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}\left[\equiv \frac{0}{0}\right]\\ =\underset{x\to 1}{\lim }\frac{\frac{1}{x^2}}{\frac{2}{x^3}}\\ =\underset{x\to 1}{\lim }\frac{1}{2}x=\frac{1}{2}\end{array}$

Therefore, $\underset{x\to 1}{\lim }\frac{1-\frac{1}{x}}{1-\frac{1}{x^2}}=\frac{1}{2}$