Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 29E from Chapter 2.6 from Stewart's Calculus, 8th Edition.
Problem 29E
Chapter:
Problem:
Find the limit or show that it does not exist.
Step-by-Step Solution
Given information
Given limit:\[L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right)\]
Step 1: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} \times \dfrac{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{{\left( {\sqrt {{x^2} + ax} } \right)}^2} - {{\left( {\sqrt {{x^2} + bx} } \right)}^2}}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + ax - {x^2} - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{ax - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\end{array}\]
Step 2: Divide Numerator and Denominator by $x$
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{ax - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\dfrac{{ax - bx}}{x}}}{{\dfrac{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}{x}}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{a - b}}{{\sqrt {\dfrac{{{x^2} + ax}}{{{x^2}}}} + \sqrt {\dfrac{{{x^2} + bx}}{{{x^2}}}} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{a - b}}{{\sqrt {1 + \dfrac{a}{x}} + \sqrt {1 + \dfrac{b}{x}} }}} \right)\end{array}\]
Step 3: Apply the Limit
\[\begin{array}{l}L = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {a - b} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {1 + \dfrac{a}{x}} + \sqrt {1 + \dfrac{b}{x}} } \right)}}} \right)\\ = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {a - b} \right)}}{{\left( {\sqrt {\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \dfrac{a}{x}} + \sqrt {\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \dfrac{b}{x}} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt {1 + \dfrac{a}{\infty }} + \sqrt {1 + \dfrac{b}{\infty }} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt {1 + 0} + \sqrt {1 + 0} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt 1 + \sqrt 1 } \right)}}} \right)\\ = \dfrac{{a - b}}{2}\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right) = \dfrac{{a - b}}{2}\]
Given limit:\[L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right)\]
Step 1: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} \times \dfrac{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{{\left( {\sqrt {{x^2} + ax} } \right)}^2} - {{\left( {\sqrt {{x^2} + bx} } \right)}^2}}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2} + ax - {x^2} - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{ax - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\end{array}\]
Step 2: Divide Numerator and Denominator by $x$
\[\begin{array}{l}L = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{ax - bx}}{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\dfrac{{ax - bx}}{x}}}{{\dfrac{{\sqrt {{x^2} + ax} + \sqrt {{x^2} + bx} }}{x}}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{a - b}}{{\sqrt {\dfrac{{{x^2} + ax}}{{{x^2}}}} + \sqrt {\dfrac{{{x^2} + bx}}{{{x^2}}}} }}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{a - b}}{{\sqrt {1 + \dfrac{a}{x}} + \sqrt {1 + \dfrac{b}{x}} }}} \right)\end{array}\]
Step 3: Apply the Limit
\[\begin{array}{l}L = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {a - b} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {1 + \dfrac{a}{x}} + \sqrt {1 + \dfrac{b}{x}} } \right)}}} \right)\\ = \left( {\dfrac{{\mathop {\lim }\limits_{x \to \infty } \left( {a - b} \right)}}{{\left( {\sqrt {\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \dfrac{a}{x}} + \sqrt {\mathop {\lim }\limits_{x \to \infty } 1 + \mathop {\lim }\limits_{x \to \infty } \dfrac{b}{x}} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt {1 + \dfrac{a}{\infty }} + \sqrt {1 + \dfrac{b}{\infty }} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt {1 + 0} + \sqrt {1 + 0} } \right)}}} \right)\\ = \left( {\dfrac{{a - b}}{{\left( {\sqrt 1 + \sqrt 1 } \right)}}} \right)\\ = \dfrac{{a - b}}{2}\end{array}\]Therefore, the Limit is \[\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + ax} - \sqrt {{x^2} + bx} } \right) = \dfrac{{a - b}}{2}\]