Explain what it means to say that lim f(x) = 9 and lim f(x) = 5. x → 1+ %3D X→ 1 O As x approaches 1, f(x) approaches 9, but f(1) = 5. O As x approaches 1 from the left, f(x) approaches 9. As x approaches 1 from the right, f(x) approaches 5. O As x approaches 1, f(x) approaches 5, but f(1) = 9. O As x approaches 1 from the right, f(x) approaches 9. As x approaches 1 from the left, f(x) approaches 5. In this situation is it possible that lim f(x) exists? Explain. x → 1 O Yes, f(x) could have a hole at (1, 9) and be defined such that f(1) = 5. O Yes, f(x) could have a hole at (1, 5) and be defined such that f(1) = 9. Yes, if f(x) has a vertical asymptote at x = 1, it can be defined such that lim f(x) = 9, lim f(x) = 5, and lim f(x) exists. x→1 メ→1+ O No, lim f(x) cannot exist if lim f(x) # lim f(x). X→1 X→1+

Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
Chapter6: Topics In Analytic Geometry
Section: Chapter Questions
Problem 33CT
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Explain what it means to say that
lim f(x) = 9 and
lim f(x) = 5.
x → 1+
%3D
X→ 1
O As x approaches 1, f(x) approaches 9, but f(1) = 5.
O As x approaches 1 from the left, f(x) approaches 9. As x approaches 1 from the right, f(x) approaches 5.
O As x approaches 1, f(x) approaches 5, but f(1) = 9.
O As x approaches 1 from the right, f(x) approaches 9. As x approaches 1 from the left, f(x) approaches 5.
In this situation is it possible that lim f(x) exists? Explain.
x → 1
O Yes, f(x) could have a hole at (1, 9) and be defined such that f(1) = 5.
O Yes, f(x) could have a hole at (1, 5) and be defined such that f(1) = 9.
Yes, if f(x) has a vertical asymptote at x = 1, it can be defined such that lim f(x) = 9, lim f(x) = 5, and lim f(x) exists.
x→1
メ→1+
O No, lim f(x) cannot exist if lim f(x) # lim f(x).
X→1
X→1+
Transcribed Image Text:Explain what it means to say that lim f(x) = 9 and lim f(x) = 5. x → 1+ %3D X→ 1 O As x approaches 1, f(x) approaches 9, but f(1) = 5. O As x approaches 1 from the left, f(x) approaches 9. As x approaches 1 from the right, f(x) approaches 5. O As x approaches 1, f(x) approaches 5, but f(1) = 9. O As x approaches 1 from the right, f(x) approaches 9. As x approaches 1 from the left, f(x) approaches 5. In this situation is it possible that lim f(x) exists? Explain. x → 1 O Yes, f(x) could have a hole at (1, 9) and be defined such that f(1) = 5. O Yes, f(x) could have a hole at (1, 5) and be defined such that f(1) = 9. Yes, if f(x) has a vertical asymptote at x = 1, it can be defined such that lim f(x) = 9, lim f(x) = 5, and lim f(x) exists. x→1 メ→1+ O No, lim f(x) cannot exist if lim f(x) # lim f(x). X→1 X→1+
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