Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 23E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 23E
Chapter:
Problem:
Evaluate the limit, if it exists.
Step-by-Step Solution
Given information
We are given with following Limit\[L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\]
Step 1:
Substitute the limit \[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\\\\ = \dfrac{{\dfrac{1}{3} - \dfrac{1}{3}}}{{3 - 3}}\\\\ = \dfrac{0}{0}\end{array}\]On substitution, we find that this is an indeterminate form.
Step 2:
Simplify the Numerator\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{{3 - x}}{{3x}}}}{{x - 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {3 - x} \right)}}{{\left( {x - 3} \right)3x}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - \left( {x - 3} \right)}}{{\left( {x - 3} \right)3x}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - 1}}{{3x}}\\ = \dfrac{{ - 1}}{{3\left( 3 \right)}}\\ = \dfrac{{ - 1}}{9}\end{array}\]Therefore, the limit is
\[\mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}} = \dfrac{{ - 1}}{9}\]
We are given with following Limit\[L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\]
Step 1:
Substitute the limit \[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\\\\ = \dfrac{{\dfrac{1}{3} - \dfrac{1}{3}}}{{3 - 3}}\\\\ = \dfrac{0}{0}\end{array}\]On substitution, we find that this is an indeterminate form.
Step 2:
Simplify the Numerator\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{{3 - x}}{{3x}}}}{{x - 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\left( {3 - x} \right)}}{{\left( {x - 3} \right)3x}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - \left( {x - 3} \right)}}{{\left( {x - 3} \right)3x}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{ - 1}}{{3x}}\\ = \dfrac{{ - 1}}{{3\left( 3 \right)}}\\ = \dfrac{{ - 1}}{9}\end{array}\]Therefore, the limit is
\[\mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{1}{x} - \dfrac{1}{3}}}{{x - 3}} = \dfrac{{ - 1}}{9}\]