10. (4 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither. 2x+12 lim x-6x+6| Solution First, work on the right-sided limit: Suppose x-6+. Then x>-6, so x+6>0. Therefore, |x+6|=x+6 (from the definition of the absolute value function). Hence, lim = 2x+12 lim 2(x+6) = lim 2=2 x-6x+6x-6+ x+6 x-6+ Reasoning: (0, 0.5, 1) Answer for the right-sided Limit: (0, 0.5) Next, work on the left-sided limit. Suppose x-6. Then x<-6, so x+6<0. Therefore, |x+6|==(x+6) (from the definition of the absolute value function). Hence, lim 2x+12 lim 2(x+6) Reasoning: (0, 0.5, 1) lim (-2)--2 x-6x+6x-6¯ −(x+6) x→−6¯ Answer for the left-sided Limit: (0,0.5) Conclusion: (0, 0.5, 1) Since the two one-sided limits are different, we conclude that the limit, lim 2x+12 does not exist x-6x+6|' and is neither co nor-00.

10. (4 pts) Evaluate the limit if it exists. If it does not exist, explain whether it is co, -co, or neither. 2x+12 lim x-6x+6| Solution First, work on the right-sided limit: Suppose x-6+. Then x>-6, so x+6>0. Therefore, |x+6|=x+6 (from the definition of the absolute value function). Hence, lim = 2x+12 lim 2(x+6) = lim 2=2 x-6x+6x-6+ x+6 x-6+ Reasoning: (0, 0.5, 1) Answer for the right-sided Limit: (0, 0.5) Next, work on the left-sided limit. Suppose x-6. Then x<-6, so x+6<0. Therefore, |x+6|==(x+6) (from the definition of the absolute value function). Hence, lim 2x+12 lim 2(x+6) Reasoning: (0, 0.5, 1) lim (-2)--2 x-6x+6x-6¯ −(x+6) x→−6¯ Answer for the left-sided Limit: (0,0.5) Conclusion: (0, 0.5, 1) Since the two one-sided limits are different, we conclude that the limit, lim 2x+12 does not exist x-6x+6|' and is neither co nor-00.

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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