Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 15E from Chapter 2.3 from Stewart's Calculus, 8th Edition.
Problem 15E
Chapter:
Problem:
Evaluate the limit, if it exists.
Step-by-Step Solution
Step 1
We are given a function whose limit we have to evaluate:\[\mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\]
Step 2:
Substitute the value of $t$ to find the function value: \[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\\ = \dfrac{{{{\left( { - 3} \right)}^2} - 9}}{{2{{\left( { - 3} \right)}^2} + 7\left( { - 3} \right) + 3}}\\ = \dfrac{{9 - 9}}{{18 - 21 + 3}}\\ = \dfrac{0}{0}\end{array}\]This forms an indeterminate form of type 0/0
Step 3:
Factor the denominator and numerator
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{2{t^2} + 6t + t + 3}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{2t\left( {t + 3} \right) + 1\left( {t + 3} \right)}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{\left( {t + 3} \right)\left( {2t + 1} \right)}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t - 3} \right)}}{{\left( {2t + 1} \right)}}\end{array}\]
Step 4: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t - 3} \right)}}{{\left( {2t + 1} \right)}}\\ = \dfrac{{\left( { - 3 - 3} \right)}}{{\left( {2\left( { - 3} \right) + 1} \right)}}\\ = \dfrac{{ - 6}}{{ - 6 + 1}}\\ = - \dfrac{6}{{ - 5}}\\ = \dfrac{6}{5}\end{array}\]Therefore,
\[\mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}} = \dfrac{6}{5}\]
We are given a function whose limit we have to evaluate:\[\mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\]
Step 2:
Substitute the value of $t$ to find the function value: \[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\\ = \dfrac{{{{\left( { - 3} \right)}^2} - 9}}{{2{{\left( { - 3} \right)}^2} + 7\left( { - 3} \right) + 3}}\\ = \dfrac{{9 - 9}}{{18 - 21 + 3}}\\ = \dfrac{0}{0}\end{array}\]This forms an indeterminate form of type 0/0
Step 3:
Factor the denominator and numerator
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{2{t^2} + 6t + t + 3}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{2t\left( {t + 3} \right) + 1\left( {t + 3} \right)}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t + 3} \right)\left( {t - 3} \right)}}{{\left( {t + 3} \right)\left( {2t + 1} \right)}}\\ = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t - 3} \right)}}{{\left( {2t + 1} \right)}}\end{array}\]
Step 4: Apply the Limit
\[\begin{array}{l}L = \mathop {\lim }\limits_{t \to - 3} \dfrac{{\left( {t - 3} \right)}}{{\left( {2t + 1} \right)}}\\ = \dfrac{{\left( { - 3 - 3} \right)}}{{\left( {2\left( { - 3} \right) + 1} \right)}}\\ = \dfrac{{ - 6}}{{ - 6 + 1}}\\ = - \dfrac{6}{{ - 5}}\\ = \dfrac{6}{5}\end{array}\]Therefore,
\[\mathop {\lim }\limits_{t \to - 3} \dfrac{{{t^2} - 9}}{{2{t^2} + 7t + 3}} = \dfrac{6}{5}\]