Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 42E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 42E
Chapter:
Problem:
Find the limit, if it exists. If the limit does not exist, explain why.
Step-by-Step Solution
Given information
We have to find the following Limit\[\mathop {\lim }\limits_{x \to - 6} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}}\]
Step 1: Left Hand Limit
Replace the Absolute function for x<6\[\begin{array}{l}\mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = \mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{ - \left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2\left( {x + 6} \right)}}{{ - \left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ - }} \left( { - 2} \right)\\ = - 2\end{array}\]Therefore, Left hand Limit is \[\mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = - 2\]
Step 2: Right Hand Limit
Replace the Absolute function for x>6\[\begin{array}{l}\mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = \mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2\left( {x + 6} \right)}}{{\left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ + }} \left( 2 \right)\\ = 2\end{array}\]Therefore, Right hand Limit is \[\mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = 2\]
Step 3: The Limit
Since Left hand side limit is not same as right hand side limit, the limit at x=6 does not exist. \[\mathop {\lim }\limits_{x \to - 6} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = {\rm{DNE}}\]
We have to find the following Limit\[\mathop {\lim }\limits_{x \to - 6} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}}\]
Step 1: Left Hand Limit
Replace the Absolute function for x<6\[\begin{array}{l}\mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = \mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{ - \left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2\left( {x + 6} \right)}}{{ - \left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ - }} \left( { - 2} \right)\\ = - 2\end{array}\]Therefore, Left hand Limit is \[\mathop {\lim }\limits_{x \to - {6^ - }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = - 2\]
Step 2: Right Hand Limit
Replace the Absolute function for x>6\[\begin{array}{l}\mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = \mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2\left( {x + 6} \right)}}{{\left( {x + 6} \right)}}\\ = \mathop {\lim }\limits_{x \to - {6^ + }} \left( 2 \right)\\ = 2\end{array}\]Therefore, Right hand Limit is \[\mathop {\lim }\limits_{x \to - {6^ + }} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = 2\]
Step 3: The Limit
Since Left hand side limit is not same as right hand side limit, the limit at x=6 does not exist. \[\mathop {\lim }\limits_{x \to - 6} \dfrac{{2x + 12}}{{\left| {x + 6} \right|}} = {\rm{DNE}}\]