Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 30E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 30E
Chapter:
Problem:
Evaluate the limit, if it exists.
Step-by-Step Solution
Given information
We are given with the limit \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right)\]
Step 1: Simplify the expression
Rationalize the numerator\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}} \times \dfrac{{\sqrt {{x^2} + 9} + 5}}{{\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} + 9 - {5^2}}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\]
Step 2: Factorize the numerator.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right)\]
Step 3: Apply the Limit.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right) = \left( {\dfrac{{\left( { - 4 - 4} \right)}}{{\sqrt {{{\left( { - 4} \right)}^2} + 9} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{\sqrt {16 + 9} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{\sqrt {25} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{5 + 5}}} \right)\\ = - \dfrac{4}{5}\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = - \dfrac{4}{5}\]
We are given with the limit \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right)\]
Step 1: Simplify the expression
Rationalize the numerator\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}} \times \dfrac{{\sqrt {{x^2} + 9} + 5}}{{\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} + 9 - {5^2}}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\]
Step 2: Factorize the numerator.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x + 4} \right)\left( {x - 4} \right)}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right)\\ = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{{x^2} - 16}}{{\left( {x + 4} \right)\sqrt {{x^2} + 9} + 5}}} \right) = \mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right)\]
Step 3: Apply the Limit.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\left( {x - 4} \right)}}{{\sqrt {{x^2} + 9} + 5}}} \right) = \left( {\dfrac{{\left( { - 4 - 4} \right)}}{{\sqrt {{{\left( { - 4} \right)}^2} + 9} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{\sqrt {16 + 9} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{\sqrt {25} + 5}}} \right)\\ = \left( {\dfrac{{\left( { - 8} \right)}}{{5 + 5}}} \right)\\ = - \dfrac{4}{5}\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to - 4} \left( {\dfrac{{\sqrt {{x^2} + 9} - 5}}{{x + 4}}} \right) = - \dfrac{4}{5}\]