Apply L'hopitals rule to evaluate the limit. Select the appropriate set of steps. lim x+2 x²+x-6 x²-4 Original limit is the form 0/0. x²+x-6 lim ( *² + *- *- - (²*xx *) - ÷ (22-0)- = = x²-4 lim x+2\ 2x 2 Original limit is the form ∞/∞ (x² + J -6 x²-4 lim x+2 lim x+2\ 2x+1 2x 5 == 4 O Original limit is the form ∞0/0⁰ 2x-6 Tim (1²+x-6) - lim (2x = 0 ) - = = = X-2 2 x-2
Apply L'hopitals rule to evaluate the limit. Select the appropriate set of steps. lim x+2 x²+x-6 x²-4 Original limit is the form 0/0. x²+x-6 lim ( *² + *- *- - (²*xx *) - ÷ (22-0)- = = x²-4 lim x+2\ 2x 2 Original limit is the form ∞/∞ (x² + J -6 x²-4 lim x+2 lim x+2\ 2x+1 2x 5 == 4 O Original limit is the form ∞0/0⁰ 2x-6 Tim (1²+x-6) - lim (2x = 0 ) - = = = X-2 2 x-2
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter1: Functions
Section1.5: Polynomial And Rational Functions
Problem 41E
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Can someone help me solve this problem with all steps included than you.
![Apply L'hopitals rule to evaluate the limit. Select the appropriate set of steps.
lim
x²+x-6
x²-4
Original limit is the form 0/0.
(x²+x-6)
2
x²-4
lim
x+2
lim
x+2
Original limit is the form ∞/⁰⁰
lim
x+2
x²+x- -6
lim
Original limit is the form ∞/⁰⁰
(x²+x-6
x²-4
x+2
2-4
- lim
X-2
2x 1 5
-9-) - m (²x + ¹) - ²
==
= lim
x-2
2x
4
2x-6
2x
x²+x- - 6
= lim
Original limit is the form 0/0.
x²-4
x-2
2x-6
2x
2
2
=
1 5
0 ) - lim ( 2x + ¹) - ²
=
==
2x
4
x+2\
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F30fc4209-43ac-409b-8363-1a56de792bf2%2F044a2e08-98af-4dc5-a0e7-4b589bfb71d0%2Faxjz2qo_processed.png&w=3840&q=75)
Transcribed Image Text:Apply L'hopitals rule to evaluate the limit. Select the appropriate set of steps.
lim
x²+x-6
x²-4
Original limit is the form 0/0.
(x²+x-6)
2
x²-4
lim
x+2
lim
x+2
Original limit is the form ∞/⁰⁰
lim
x+2
x²+x- -6
lim
Original limit is the form ∞/⁰⁰
(x²+x-6
x²-4
x+2
2-4
- lim
X-2
2x 1 5
-9-) - m (²x + ¹) - ²
==
= lim
x-2
2x
4
2x-6
2x
x²+x- - 6
= lim
Original limit is the form 0/0.
x²-4
x-2
2x-6
2x
2
2
=
1 5
0 ) - lim ( 2x + ¹) - ²
=
==
2x
4
x+2\
=
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