Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 26E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 26E
Chapter:
Problem:
Evaluate the limit, if it exists.
Step-by-Step Solution
Given information
We are given with the limit \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right)\]
Step 1: Simplify the expression
\[\begin{array}{l}\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right) = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{t + 1 - 1}}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{t}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right) = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right)\]
Step 2: Apply limit in numerator and denominator.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right) = \dfrac{{\mathop {\lim }\limits_{t \to 0} 1}}{{\mathop {\lim }\limits_{t \to 0} \left( {t + 1} \right)}}\\ = \dfrac{{\mathop {\lim }\limits_{t \to 0} 1}}{{\mathop {\lim }\limits_{t \to 0} \left( t \right) + \mathop {\lim }\limits_{t \to 0} \left( 1 \right)}}\\ = \dfrac{1}{{0 + 1}}\\ = 1\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right) = 1\]
We are given with the limit \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right)\]
Step 1: Simplify the expression
\[\begin{array}{l}\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right) = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{t + 1 - 1}}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{t}{{t\left( {t + 1} \right)}}} \right)\\ = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right)\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2} + t}}} \right) = \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right)\]
Step 2: Apply limit in numerator and denominator.
\[\begin{array}{l}\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right) = \dfrac{{\mathop {\lim }\limits_{t \to 0} 1}}{{\mathop {\lim }\limits_{t \to 0} \left( {t + 1} \right)}}\\ = \dfrac{{\mathop {\lim }\limits_{t \to 0} 1}}{{\mathop {\lim }\limits_{t \to 0} \left( t \right) + \mathop {\lim }\limits_{t \to 0} \left( 1 \right)}}\\ = \dfrac{1}{{0 + 1}}\\ = 1\end{array}\]Therefore, \[\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{1}{{\left( {t + 1} \right)}}} \right) = 1\]