(3) Using the appropriate approach covered in Unit 6 of this Course along with knowledge gained in Unit 1 and 2, show that: (a) Limx→0 (1/x-1 +1/x+1)/x = -2 (b) Limh→0 (h/√5h + 4 − 2) = 4/5 (c) Limt→1(t3- 1/t-1) = 3

(3) Using the appropriate approach covered in Unit 6 of this Course along with knowledge gained in Unit 1 and 2, show that: (a) Limx→0 (1/x-1 +1/x+1)/x = -2 (b) Limh→0 (h/√5h + 4 − 2) = 4/5 (c) Limt→1(t3- 1/t-1) = 3

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.

Chapter4: Calculating The Derivative

Section4.1: Techniques For Finding Derivatives

Problem 35E

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Concept explainers

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.

Question

(3) Using the appropriate approach covered in Unit 6 of this Course along with knowledge
gained in Unit 1 and 2, show that:

(a) Lim_x→0(1/x-1 +1/x+1)/x = -2

(b) Lim_h→0(h/√5h + 4 − 2) = 4/5

(c) Lim_t→1(t³- 1/t-1) = 3

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