University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises: 31

Answer

$\lim\limits_{x \to 1}\frac{x^{-1}-1}{(x-1)} = -1$

Work Step by Step

Simplify: $\frac{x^{-1}-1}{(x-1)} = \frac{(1-x)}{x(x-1)} = -\frac{(x-1)}{x(x-1)} = -\frac{1}{x}$ Now, $\lim\limits_{x \to 1}\frac{x^{-1}-1}{(x-1)} = -\lim\limits_{x \to 1}\frac{1}{x} = -1$

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