Calculus: Early Transcendentals, AP Edition, 6th Edition by James Stewart and a great selection of related books, art and collectibles available now at AbeBooks.com. We are proud to announce the author team who will continue the best-selling James Stewart Calculus franchise. Saleem Watson, who received his doctorate degree under Stewart’s instruction, and Daniel Clegg, a former colleague of Stewart’s, will author the revised series, which has been used by more than 8 million students over the last fifteen years.
We have solutions for your book!
See our solution for Question 10E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Chapter:
What is wrong with the following equation?
Given information
We are given with following Inequality
[dfrac{{{x^2} + x - 6}}{{x - 2}} = x + 3]
Step 1: Part (a)
First we have to find what is wrong with the inequality. Any polynomial function of the form $dfrac{{pleft( x right)}}{{qleft( x right)}}$ can only be defined if the denominator ${qleft( x right)}$ is non zero. Thus to write the polynomial $dfrac{{pleft( x right)}}{{qleft( x right)}}$, we must also write [dfrac{{pleft( x right)}}{{qleft( x right)}} = 0,left{ {qleft( x right) ne 0} right}]For the left hand side, the denominator is: ${x - 2}$, that is 0 when $x=2$, so the left hand side is not defined when $x=2$. However, the right hand side is defined even at $x=2$. So to write this inequality, we must also mention that this inequality is not defined at $x=2$. In that manner the given inequality is not correct.
Step 2: Part (b)
We are given with the limit form of the inequality: [mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} left( {x + 3} right)]We will find limits of both the sides to prove the inequality.
Step 3: Limit of Left hand side function
[begin{array}{l}mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{{x^2} - 2x + 3x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{xleft( {x - 2} right) + 3left( {x - 2} right)}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{left( {x - 2} right)left( {x + 3} right)}}{{x - 2}} = mathop {lim }limits_{x to 2} left( {x + 3} right) = left( {2 + 3} right) = 5end{array}]We can see that left hand side limit [mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = 5]
Step 4: Limit of Right hand side function
[begin{array}{l}mathop {lim }limits_{x to 2} left( {x + 3} right) = left( {2 + 3} right) = 5end{array}]We can see that Right hand side limit [mathop {lim }limits_{x to 2} left( {x + 3} right) = 5]Hence the above inequality is Correct
We are given with following Inequality
[dfrac{{{x^2} + x - 6}}{{x - 2}} = x + 3]
Step 1: Part (a)
First we have to find what is wrong with the inequality. Any polynomial function of the form $dfrac{{pleft( x right)}}{{qleft( x right)}}$ can only be defined if the denominator ${qleft( x right)}$ is non zero. Thus to write the polynomial $dfrac{{pleft( x right)}}{{qleft( x right)}}$, we must also write [dfrac{{pleft( x right)}}{{qleft( x right)}} = 0,left{ {qleft( x right) ne 0} right}]For the left hand side, the denominator is: ${x - 2}$, that is 0 when $x=2$, so the left hand side is not defined when $x=2$. However, the right hand side is defined even at $x=2$. So to write this inequality, we must also mention that this inequality is not defined at $x=2$. In that manner the given inequality is not correct.
Step 2: Part (b)
We are given with the limit form of the inequality: [mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} left( {x + 3} right)]We will find limits of both the sides to prove the inequality.
Step 3: Limit of Left hand side function
[begin{array}{l}mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{{x^2} - 2x + 3x - 6}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{xleft( {x - 2} right) + 3left( {x - 2} right)}}{{x - 2}} = mathop {lim }limits_{x to 2} dfrac{{left( {x - 2} right)left( {x + 3} right)}}{{x - 2}} = mathop {lim }limits_{x to 2} left( {x + 3} right) = left( {2 + 3} right) = 5end{array}]We can see that left hand side limit [mathop {lim }limits_{x to 2} dfrac{{{x^2} + x - 6}}{{x - 2}} = 5]
Step 4: Limit of Right hand side function
[begin{array}{l}mathop {lim }limits_{x to 2} left( {x + 3} right) = left( {2 + 3} right) = 5end{array}]We can see that Right hand side limit [mathop {lim }limits_{x to 2} left( {x + 3} right) = 5]Hence the above inequality is Correct
Step 1 of 2
Quisque eget sagittis purus. Nunc sagittis nisi magna, in mollis lectus ullamcorper in. Sed sodales risus sed arcu efficitur, id rutrum ligula laoreet. Quisque molestie purus sed consequat fermentum. Fusce ut lectus lobortis, viverra sem nec, rhoncus justo. Phasellus malesuada, ipsum ac varius euismod, purus nulla volutpat nunc, eu fermentum odio justo porttitor libero. Quisque viverra arcu nibh, at facilisis tortor ornare non. Etiam id porttitor arcu, ut eleifend nisi. Ut sit amet enim eu lacus egestas tristique eleifend sit amet lectus.
Step 2 of 2
Nam consectetur iaculis dui ac tempor. Nulla a nisi nunc. Suspendisse semper mauris pretium, suscipit sapien nec, hendrerit justo. In hac habitasse platea dictumst. Praesent laoreet gravida posuere. Maecenas interdum ante nec libero pellentesque, sit amet commodo nisl auctor. Phasellus facilisis, lorem et fringilla varius, mi felis rutrum diam, quis ultricies mauris nisl nec nisl. Fusce lacinia tincidunt urna sit amet vehicula. Proin sed dui vitae nisi vehicula imperdiet eu a lorem. Praesent at ante nibh. Quisque id elit ac purus vestibulum auctor. Etiam ac tincidunt velit. Aenean accumsan risus tempor tincidunt luctus.