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Problem 2. Define a function h(x) on (-∞, +∞) by the following rule, sin (1/x), x 0. x = 0 h(x) = { 0, Compute the limit as x approaches 0 of k(x) = xh(x). Explain why it is invalid to simply compute the limit as k(0) = 0.h(0)=0.0=0?

Problem 2. Define a function h(x) on (-∞, +∞) by the following rule, sin (1/x), x 0. x = 0 h(x) = { 0, Compute the limit as x approaches 0 of k(x) = xh(x). Explain why it is invalid to simply compute the limit as k(0) = 0.h(0)=0.0=0?

Calculus For The Life Sciences
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
Section4.2: Derivatives Of Products And Quotients
Problem 36E
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Problem 1. The following function has domain (-1,0) U (0, ∞0), −1+√1+x Xx f(x) = Make a table of values of f(x) for x = ±1, 0.1, 0.01, and ±0.001. Use your table to guess a limiting value for f(x) as x → 0. Finally, use difference- of-squares rationalization to find a continuous function g(x) whose domain contains (-1, ∞) such that f(x) equals g(x) on (-1,0) U (0, ∞o). Use this to prove that your guess is correct. Problem 2. Define a function h(x) on (-∞, +∞) by the following rule, sin (1/x), x 0. x = 0 h(x) =X{ 0, Compute the limit as x approaches 0 of k(x) = xh(x). Explain why it is invalid to simply compute the limit as k(0) = 0.h(0)=0.0=0? 81 YOUKY
Transcribed Image Text:Problem 1. The following function has domain (-1,0) U (0, ∞0), −1+√1+x Xx f(x) = Make a table of values of f(x) for x = ±1, 0.1, 0.01, and ±0.001. Use your table to guess a limiting value for f(x) as x → 0. Finally, use difference- of-squares rationalization to find a continuous function g(x) whose domain contains (-1, ∞) such that f(x) equals g(x) on (-1,0) U (0, ∞o). Use this to prove that your guess is correct. Problem 2. Define a function h(x) on (-∞, +∞) by the following rule, sin (1/x), x 0. x = 0 h(x) =X{ 0, Compute the limit as x approaches 0 of k(x) = xh(x). Explain why it is invalid to simply compute the limit as k(0) = 0.h(0)=0.0=0? 81 YOUKY
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