Explanation Given information: A limit function is given as, lim x → 0 x + sin 3 x x 2 + 2 x . Concept used: L'Hospital's Rule: If f and g are differentiable in an open interval containing a and f ( a ) = g ( a ) = 0 . Also, g ' ( x ) ≠ 0 (except possibly at a ). Then- lim x → a f ( x ) g ( x ) = lim x → a f ' ( x ) g ' ( x ) This rule is also applicable when f ( a ) = ± ∞ and g ( a ) = ± ∞ . Calculation: Given limit function is, lim x → 0 x + sin 3 x x 2 + 2 x . Let us suppose, f ( x ) = x + sin 3 x and g ( x ) = x 2 + 2 x . Here, a = 0 . So, f ( a ) = 0 + sin 3 0 = 0 and g ( a ) = 0 2 + 2 ⋅ 0 Therefore, limit converts into an indeterminate form of 0 0

6th Edition

ISBN: 9781465208880

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 4.5, Problem 15PS

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To calculate: The value of given limit.

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Chapter 4.5, Problem 14PS

Chapter 4.5, Problem 16PS

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Chapter 4 Solutions

Calculus

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Definite Integral Calculus Examples, Integration - Basic Introduction, Practice Problems; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=rCWOdfQ3cwQ;License: Standard YouTube License, CC-BY

The value of given limit.

Solution Summary: The author explains how L'Hospital's Rule is applicable when f(x)=x+mathrmsin

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To calculate: The value of given limit.

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Chapter 4 Solutions