Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 18E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 18E
Chapter:
Problem:
Evaluate the limit, if it exists.
Step-by-Step Solution
Given information
Given limit:\[L = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - 8}}{h}} \right)\]
Step 1: Expand the numerator
Use the formula: \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - {2^3}}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {2 + h - 2} \right)\left[ {{{\left( {2 + h} \right)}^2} + \left( {2 + h} \right) \times 2 + {2^2}} \right]}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( h \right)\left[ {{{\left( {2 + h} \right)}^2} + 4 + 2h + 4} \right]}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\end{array}\]
Step 2: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\\ = \left( {\dfrac{{\left[ {{{\left( {2 + 0} \right)}^2} + 8 + 2\left( 0 \right)} \right]}}{1}} \right)\\ = \left( {\dfrac{{\left[ {4 + 8} \right]}}{1}} \right)\\ = 12\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - 8}}{h}} \right) = 12\]
Given limit:\[L = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - 8}}{h}} \right)\]
Step 1: Expand the numerator
Use the formula: \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\]\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - {2^3}}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( {2 + h - 2} \right)\left[ {{{\left( {2 + h} \right)}^2} + \left( {2 + h} \right) \times 2 + {2^2}} \right]}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left( h \right)\left[ {{{\left( {2 + h} \right)}^2} + 4 + 2h + 4} \right]}}{h}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\\ = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\end{array}\]
Step 2: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\left[ {{{\left( {2 + h} \right)}^2} + 8 + 2h} \right]}}{1}} \right)\\ = \left( {\dfrac{{\left[ {{{\left( {2 + 0} \right)}^2} + 8 + 2\left( 0 \right)} \right]}}{1}} \right)\\ = \left( {\dfrac{{\left[ {4 + 8} \right]}}{1}} \right)\\ = 12\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{{{\left( {2 + h} \right)}^3} - 8}}{h}} \right) = 12\]