Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 46E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 46E
Chapter:
Problem:
Find the limit, if it exists. If the limit does not exist, explain why.
Step-by-Step Solution
Given information
We are given with following relationship:\[L = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right)\]We have to find the right hand side of the limit.
Step 1:
The absolute function, is given by; \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}x&{x < 0}\\{ - x}&{x < 0}\end{array}} \right.\]Since we are considering function from rights side, the x value is positive. So we can replace the absolute function as $\left| x \right| = x$
Step 2: The Limit
\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{x}} \right)\\ = \mathop {\lim }\limits_{x \to {0^ + }} \left( 0 \right)\\ = 0\end{array}\]Therefore, \[\mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right) = 0\]
We are given with following relationship:\[L = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right)\]We have to find the right hand side of the limit.
Step 1:
The absolute function, is given by; \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}x&{x < 0}\\{ - x}&{x < 0}\end{array}} \right.\]Since we are considering function from rights side, the x value is positive. So we can replace the absolute function as $\left| x \right| = x$
Step 2: The Limit
\[\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{x}} \right)\\ = \mathop {\lim }\limits_{x \to {0^ + }} \left( 0 \right)\\ = 0\end{array}\]Therefore, \[\mathop {\lim }\limits_{x \to {0^ + }} \left( {\dfrac{1}{x} - \dfrac{1}{{\left| x \right|}}} \right) = 0\]