Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 10E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 10E
Chapter:
Problem:
What is wrong with the following equation?
Step-by-Step Solution
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 $\dfrac{{p\left( x \right)}}{{q\left( x \right)}}$ can only be defined if the denominator ${q\left( x \right)}$ is non zero. Thus to write the polynomial $\dfrac{{p\left( x \right)}}{{q\left( x \right)}}$, we must also write \[\dfrac{{p\left( x \right)}}{{q\left( x \right)}} = 0\,\,\,\left\{ {q\left( x \right) \ne 0} \right\}\]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
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} + x - 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 2x + 3x - 6}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{x\left( {x - 2} \right) + 3\left( {x - 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \left( {x + 3} \right)\\ = \left( {2 + 3} \right)\\ = 5\end{array}\]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
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {x + 3} \right) = \left( {2 + 3} \right)\\ = 5\end{array}\]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 $\dfrac{{p\left( x \right)}}{{q\left( x \right)}}$ can only be defined if the denominator ${q\left( x \right)}$ is non zero. Thus to write the polynomial $\dfrac{{p\left( x \right)}}{{q\left( x \right)}}$, we must also write \[\dfrac{{p\left( x \right)}}{{q\left( x \right)}} = 0\,\,\,\left\{ {q\left( x \right) \ne 0} \right\}\]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
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} + x - 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 2x + 3x - 6}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{x\left( {x - 2} \right) + 3\left( {x - 2} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{x - 2}}\\ = \mathop {\lim }\limits_{x \to 2} \left( {x + 3} \right)\\ = \left( {2 + 3} \right)\\ = 5\end{array}\]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
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {x + 3} \right) = \left( {2 + 3} \right)\\ = 5\end{array}\]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