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B.) For numbers 1 to 10: Determine if limit of each function exist using the piece-wise function below x - 3 if x < -2 if -2 3 1.) lim f(x) = 2.) lim f(x) = X-3 x→0 3.) lim f(x) = 3.) lim f(x) = x--2 x-1 5.) lim f(x) = 6.) f (0)= x→0+ 7.) lim f(x) = 8.) f(-2)= x-2- 9.) lim f(x) = 10.) f (4)= X-3+ 4x + 5 2x² - 1

B.) For numbers 1 to 10: Determine if limit of each function exist using the piece-wise function below x - 3 if x < -2 if -2 3 1.) lim f(x) = 2.) lim f(x) = X-3 x→0 3.) lim f(x) = 3.) lim f(x) = x--2 x-1 5.) lim f(x) = 6.) f (0)= x→0+ 7.) lim f(x) = 8.) f(-2)= x-2- 9.) lim f(x) = 10.) f (4)= X-3+ 4x + 5 2x² - 1

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
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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Show step-by-step solution. Please answer numbers 5-10.
B.) For numbers 1 to 10: Determine if limit of each function exist using the piece-wise function below x - 3 if x < -2 if -2<x<0 x² + 1 F(x) = if 0 < x≤ 3 if x > 3 1.) lim f(x) = 2.) lim f(x) = X-3 x→0 3.) lim f(x) = 3.) lim f(x) = x--2 x-1 5.) lim f(x) = 6.) f (0)= x→0+ 7.) lim f(x) = 8.) f(-2)= x-2- 9.) lim f(x) = 10.) f (4)= X-3+ 4x + 5 2x² - 1
Transcribed Image Text:B.) For numbers 1 to 10: Determine if limit of each function exist using the piece-wise function below x - 3 if x < -2 if -2<x<0 x² + 1 F(x) = if 0 < x≤ 3 if x > 3 1.) lim f(x) = 2.) lim f(x) = X-3 x→0 3.) lim f(x) = 3.) lim f(x) = x--2 x-1 5.) lim f(x) = 6.) f (0)= x→0+ 7.) lim f(x) = 8.) f(-2)= x-2- 9.) lim f(x) = 10.) f (4)= X-3+ 4x + 5 2x² - 1
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