Theorems on limits 3.17. If lim f(x) exists, prove that it must be unique. We must show that if lim f(x) = l, and lim f(x) = l½, then l = 12. x→x, By hypothesis, given any e >0 we can find 8 > 0 such that |f(x) – 14 | < e/2 0< |x- xol < ô when \fx) – 14|
Theorems on limits 3.17. If lim f(x) exists, prove that it must be unique. We must show that if lim f(x) = l, and lim f(x) = l½, then l = 12. x→x, By hypothesis, given any e >0 we can find 8 > 0 such that |f(x) – 14 | < e/2 0< |x- xol < ô when \fx) – 14|
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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3.17) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
![Theorems on limits
3.17.
If lim f(x) exists, prove that it must be unique.
We must show that if lim f(x) = l, and lim f(x) = l½, then l = l,.
x→x,
By hypothesis, given any e >0 we can find 8 > 0 such that
|f(x) – 14 | < e/2
0< ]x- xo] < 8
when
\f(x) – 1½] <e
< €/2
0 < |x-xol < 8
when
CHAPTER 3 Functions, Limits, and Continuity
59
Then by the absolute value property 2 on Page 4,
14 –14| = |4 -f(x) +f(x) – lz| < I4-f(x)|+ \f(x) – 1½l< e/2 + •
€/2 = €
i.e., |4-2| is less than any positive number e (however small) and so must be zero. Thus, l = l½.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2Fd7ea2f54-a8e3-414f-93bc-30a1ca63af90%2F3mzb90c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorems on limits
3.17.
If lim f(x) exists, prove that it must be unique.
We must show that if lim f(x) = l, and lim f(x) = l½, then l = l,.
x→x,
By hypothesis, given any e >0 we can find 8 > 0 such that
|f(x) – 14 | < e/2
0< ]x- xo] < 8
when
\f(x) – 1½] <e
< €/2
0 < |x-xol < 8
when
CHAPTER 3 Functions, Limits, and Continuity
59
Then by the absolute value property 2 on Page 4,
14 –14| = |4 -f(x) +f(x) – lz| < I4-f(x)|+ \f(x) – 1½l< e/2 + •
€/2 = €
i.e., |4-2| is less than any positive number e (however small) and so must be zero. Thus, l = l½.
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