7. (3 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither. lim 2x²+3x+1 x-x²-2x-3 Solution Direct substitution of "-1" into the expression yields the indeterminate form of "0/0". Instead, we proceed as follows by factoring the numerator and denominator and cancelling the common factor, (x+1), before taking the limit': lim 2x²+3x+1 x-x²-2x-3 = lim (0, 0.5, 1) (0, 0.5, 1) (0, 0.5, 1) 2x+1 lim 2(−1)+1 _ −1 _ 1 x-x-3 -1-3 -4 4 (2x+1)(x+1) x-1 (x-3)(x+1)
7. (3 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither. lim 2x²+3x+1 x-x²-2x-3 Solution Direct substitution of "-1" into the expression yields the indeterminate form of "0/0". Instead, we proceed as follows by factoring the numerator and denominator and cancelling the common factor, (x+1), before taking the limit': lim 2x²+3x+1 x-x²-2x-3 = lim (0, 0.5, 1) (0, 0.5, 1) (0, 0.5, 1) 2x+1 lim 2(−1)+1 _ −1 _ 1 x-x-3 -1-3 -4 4 (2x+1)(x+1) x-1 (x-3)(x+1)
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter6: Exponential Functions And Sequences
Section: Chapter Questions
Problem 7CA
Related questions
Question
Can you show the steps of how to get this answer, and explain all the steps.
![7. (3 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither.
lim
2x²+3x+1
x-x²-2x-3
Solution Direct substitution of "-1" into the expression yields the indeterminate form of "0/0".
Instead, we proceed as follows by factoring the numerator and denominator and cancelling the
common factor, (x+1), before taking the limit':
lim
2x²+3x+1
x-x²-2x-3
=
lim
(0, 0.5, 1)
(0, 0.5, 1)
(0, 0.5, 1)
2x+1
lim
2(−1)+1 _ −1 _ 1
x-x-3
-1-3 -4 4
(2x+1)(x+1)
x-1 (x-3)(x+1)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6356ceb6-8738-457b-86a1-1decb02218d2%2Fcda867b2-41a0-4936-96d4-f94f1c78ab39%2Fzwrk83_processed.png&w=3840&q=75)
Transcribed Image Text:7. (3 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither.
lim
2x²+3x+1
x-x²-2x-3
Solution Direct substitution of "-1" into the expression yields the indeterminate form of "0/0".
Instead, we proceed as follows by factoring the numerator and denominator and cancelling the
common factor, (x+1), before taking the limit':
lim
2x²+3x+1
x-x²-2x-3
=
lim
(0, 0.5, 1)
(0, 0.5, 1)
(0, 0.5, 1)
2x+1
lim
2(−1)+1 _ −1 _ 1
x-x-3
-1-3 -4 4
(2x+1)(x+1)
x-1 (x-3)(x+1)
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