Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 2E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 2E
Chapter:
Problem:
The graphs of f and t are given. Use them to evaluate each limit, if it exists. If the limit does not exist, explain why.
Step-by-Step Solution
Given information
We are given with graph of functions $f$ and $g$
Step 1: (a)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) = - 1\\\mathop {\lim }\limits_{x \to 2} \left( {g\left( x \right)} \right) = 2\end{array}\]Thus using the sum Law:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) + \mathop {\lim }\limits_{x \to2} \left( {g\left( x \right)} \right)\\ = - 1 + 2\\ = 1\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to2} \left( {f\left( x \right) + g\left( x \right)} \right) = 1\]
Step 2: (b)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right)\]From Figure we have: \[\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right)} \right) = 2\]For $g(x)$, limit does not exist at 0: However, the right hand limit is: \[\mathop {\lim }\limits_{x \to 0 + } \left( {g\left( x \right)} \right) = 3\]The left hand limit is: \[\mathop {\lim }\limits_{x \to 0 - } \left( {g\left( x \right)} \right) = 1\]Substitute to find the right hand limit of given expression: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right)} \right) - \mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right)} \right)\\ = 2 - 3\\ = - 1\end{array}\]The left hand limit of the expression:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right)} \right) - \mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right)} \right)\\ = 2 - 1\\ = 1\end{array}\]As the left hand limit and right hand limit are different, The limit does not exist:Therefore, the Limit is The limit Does not Exist
Step 3: (c)
We have to find the expression: \[\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)} \right) = 1\\\mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 2\end{array}\]Use the Product Rule:\[\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to - 1} f\left( x \right) \times \mathop {\lim }\limits_{x \to - 1} g\left( x \right)\\ = \mathop {\lim }\limits_{x \to - 1} f\left( x \right) \times \mathop {\lim }\limits_{x \to - 1} g\left( x \right)\\ = 1 \times 2\\ = 2\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right) = 2\]
Step 4: (d)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {f\left( x \right)} \right) = 1\\\mathop {\lim }\limits_{x \to 3} \left( {g\left( x \right)} \right) = 0\end{array}\]Use the Quotient Rule:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \left( {\dfrac{{\mathop {\lim }\limits_{x \to 3} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 3} g\left( x \right)}}} \right)\\ = \left( {\dfrac{1}{0}} \right)\\ = {\rm{DNE}}\end{array}\]Therefore, the Limit Does Not Exist
Step 5: (e)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right)\]From Figure we have: \[\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) = - 1\]Apply the limits using product rule\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = \left( {\mathop {\lim }\limits_{x \to 2} {x^2}\mathop {\lim }\limits_{x \to 2} f\left( x \right)} \right)\\ = \left( {{2^2}\left( { - 1} \right)} \right)\\ = - 4\end{array}\]Therefore, the Limit \[\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = - 4\]
Step 6: (f)
We have to find the expression: \[f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}f\left( { - 1} \right) = 3\\\mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 2\end{array}\]Apply the limits\[\begin{array}{l}f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 3 + 2\\ = 5\end{array}\]Therefore, the Limit \[f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 5\]
We are given with graph of functions $f$ and $g$
Step 1: (a)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) = - 1\\\mathop {\lim }\limits_{x \to 2} \left( {g\left( x \right)} \right) = 2\end{array}\]Thus using the sum Law:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) + \mathop {\lim }\limits_{x \to2} \left( {g\left( x \right)} \right)\\ = - 1 + 2\\ = 1\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to2} \left( {f\left( x \right) + g\left( x \right)} \right) = 1\]
Step 2: (b)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right) - g\left( x \right)} \right)\]From Figure we have: \[\mathop {\lim }\limits_{x \to 0} \left( {f\left( x \right)} \right) = 2\]For $g(x)$, limit does not exist at 0: However, the right hand limit is: \[\mathop {\lim }\limits_{x \to 0 + } \left( {g\left( x \right)} \right) = 3\]The left hand limit is: \[\mathop {\lim }\limits_{x \to 0 - } \left( {g\left( x \right)} \right) = 1\]Substitute to find the right hand limit of given expression: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right)} \right) - \mathop {\lim }\limits_{x \to 0 + } \left( {f\left( x \right)} \right)\\ = 2 - 3\\ = - 1\end{array}\]The left hand limit of the expression:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right)} \right) - \mathop {\lim }\limits_{x \to 0 - } \left( {f\left( x \right)} \right)\\ = 2 - 1\\ = 1\end{array}\]As the left hand limit and right hand limit are different, The limit does not exist:Therefore, the Limit is The limit Does not Exist
Step 3: (c)
We have to find the expression: \[\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)} \right) = 1\\\mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 2\end{array}\]Use the Product Rule:\[\begin{array}{l}\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to - 1} f\left( x \right) \times \mathop {\lim }\limits_{x \to - 1} g\left( x \right)\\ = \mathop {\lim }\limits_{x \to - 1} f\left( x \right) \times \mathop {\lim }\limits_{x \to - 1} g\left( x \right)\\ = 1 \times 2\\ = 2\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to - 1} \left( {f\left( x \right)g\left( x \right)} \right) = 2\]
Step 4: (d)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right)\]From Figure we have: \[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {f\left( x \right)} \right) = 1\\\mathop {\lim }\limits_{x \to 3} \left( {g\left( x \right)} \right) = 0\end{array}\]Use the Quotient Rule:\[\begin{array}{l}\mathop {\lim }\limits_{x \to 3} \left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \left( {\dfrac{{\mathop {\lim }\limits_{x \to 3} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 3} g\left( x \right)}}} \right)\\ = \left( {\dfrac{1}{0}} \right)\\ = {\rm{DNE}}\end{array}\]Therefore, the Limit Does Not Exist
Step 5: (e)
We have to find the expression: \[\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right)\]From Figure we have: \[\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right)} \right) = - 1\]Apply the limits using product rule\[\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = \left( {\mathop {\lim }\limits_{x \to 2} {x^2}\mathop {\lim }\limits_{x \to 2} f\left( x \right)} \right)\\ = \left( {{2^2}\left( { - 1} \right)} \right)\\ = - 4\end{array}\]Therefore, the Limit \[\mathop {\lim }\limits_{x \to 2} \left( {{x^2}f\left( x \right)} \right) = - 4\]
Step 6: (f)
We have to find the expression: \[f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right)\]From Figure we have: \[\begin{array}{l}f\left( { - 1} \right) = 3\\\mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 2\end{array}\]Apply the limits\[\begin{array}{l}f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 3 + 2\\ = 5\end{array}\]Therefore, the Limit \[f\left( { - 1} \right) + \mathop {\lim }\limits_{x \to - 1} \left( {g\left( x \right)} \right) = 5\]