Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 27E from Chapter 2.6 from Stewart's Calculus, 8th Edition.
Problem 27E
Chapter:
Problem:
Find the limit or show that it does not exist.
Step-by-Step Solution
Given information
Given limit:\[L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right)\]
Step 1: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x \times \dfrac{{\sqrt {9{x^2} + x} + 3x}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{{\left( {\sqrt {9{x^2} + x} } \right)}^2} - {{\left( {3x} \right)}^2}}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{9{x^2} + x - 9{x^2}}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{x}{{\sqrt {9{x^2} + x} + 3x}}} \right)\end{array}\]
Step 2: Divide Numerator and Denominator by $x$
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{x}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x/x}}{{\left( {\sqrt {9{x^2} + x} + 3x} \right)/x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {\dfrac{{9{x^2} + x}}{{{x^2}}}} + \dfrac{{3x}}{x}} \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {9 + \dfrac{1}{x}} + 3} \right)}}} \right)\end{array}\]
Step 3: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {9 + \dfrac{1}{x}} + 3} \right)}}} \right)\\ = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\left( {\sqrt {\mathop {\lim }\limits_{x \to \infty } 9 + \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{x}} \right)} + \mathop {\lim }\limits_{x \to \infty } 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {\sqrt {9 + \left( {\dfrac{1}{\infty }} \right)} + 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {\sqrt 9 + 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {3 + 3} \right)}}} \right)\\ = \dfrac{1}{6}\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right) = \dfrac{1}{6}\]
Given limit:\[L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right)\]
Step 1: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x \times \dfrac{{\sqrt {9{x^2} + x} + 3x}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{{\left( {\sqrt {9{x^2} + x} } \right)}^2} - {{\left( {3x} \right)}^2}}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{9{x^2} + x - 9{x^2}}}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{x}{{\sqrt {9{x^2} + x} + 3x}}} \right)\end{array}\]
Step 2: Divide Numerator and Denominator by $x$
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{x}{{\sqrt {9{x^2} + x} + 3x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{x/x}}{{\left( {\sqrt {9{x^2} + x} + 3x} \right)/x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {\dfrac{{9{x^2} + x}}{{{x^2}}}} + \dfrac{{3x}}{x}} \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {9 + \dfrac{1}{x}} + 3} \right)}}} \right)\end{array}\]
Step 3: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{\left( {\sqrt {9 + \dfrac{1}{x}} + 3} \right)}}} \right)\\ = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } 1}}{{\left( {\sqrt {\mathop {\lim }\limits_{x \to \infty } 9 + \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{x}} \right)} + \mathop {\lim }\limits_{x \to \infty } 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {\sqrt {9 + \left( {\dfrac{1}{\infty }} \right)} + 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {\sqrt 9 + 3} \right)}}} \right)\\ = \left( {\dfrac{1}{{\left( {3 + 3} \right)}}} \right)\\ = \dfrac{1}{6}\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {9{x^2} + x} - 3x} \right) = \dfrac{1}{6}\]