Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 29E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 29E
Chapter:
Problem:
Step-by-Step Solution
Step 1
We are given with following limit:\[L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{t\sqrt {t + 1} }} - \dfrac{1}{t}} \right)\]
Step 2:
Substitute the limit to see the form of the limit:\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{t\sqrt {t + 1} }} - \dfrac{1}{t}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }}} \right)\\ = \left( {\dfrac{{1 - \sqrt {0 + 1} }}{{\left( 0 \right)\sqrt {0 + 1} }}} \right)\\ = \left( {\dfrac{{1 - 1}}{{\left( 0 \right)\sqrt 1 }}} \right)\\ = \left( {\dfrac{0}{0}} \right)\end{array}\]This is an indeterminate form:
Step 3: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }} \times \dfrac{{1 + \sqrt {t + 1} }}{{1 + \sqrt {t + 1} }}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{1^2} - {{\left( {\sqrt {t + 1} } \right)}^2}}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \left( {t + 1} \right)}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{-t}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\end{array}\]
Step 4: Simplify and Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{ - t}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{ - 1}}{{\left( {\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{{\left( {\sqrt {0 + 1} } \right)\left( {1 + \sqrt {0 + 1} } \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{{\left( 1 \right)\left( {1 + 1} \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{2}} \right)\\ = - \dfrac{1}{2}\end{array}\]Therefore, \[L = - \dfrac{1}{2}\]
We are given with following limit:\[L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{t\sqrt {t + 1} }} - \dfrac{1}{t}} \right)\]
Step 2:
Substitute the limit to see the form of the limit:\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{t\sqrt {t + 1} }} - \dfrac{1}{t}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }}} \right)\\ = \left( {\dfrac{{1 - \sqrt {0 + 1} }}{{\left( 0 \right)\sqrt {0 + 1} }}} \right)\\ = \left( {\dfrac{{1 - 1}}{{\left( 0 \right)\sqrt 1 }}} \right)\\ = \left( {\dfrac{0}{0}} \right)\end{array}\]This is an indeterminate form:
Step 3: Rationalize the function
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \sqrt {t + 1} }}{{t\sqrt {t + 1} }} \times \dfrac{{1 + \sqrt {t + 1} }}{{1 + \sqrt {t + 1} }}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{{1^2} - {{\left( {\sqrt {t + 1} } \right)}^2}}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{1 - \left( {t + 1} \right)}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{-t}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\end{array}\]
Step 4: Simplify and Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{ - t}}{{\left( {t\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{ - 1}}{{\left( {\sqrt {t + 1} } \right)\left( {1 + \sqrt {t + 1} } \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{{\left( {\sqrt {0 + 1} } \right)\left( {1 + \sqrt {0 + 1} } \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{{\left( 1 \right)\left( {1 + 1} \right)}}} \right)\\ = \left( {\dfrac{{ - 1}}{2}} \right)\\ = - \dfrac{1}{2}\end{array}\]Therefore, \[L = - \dfrac{1}{2}\]