Explanation Given: The given limit is lim x → 0 cot 3 x cot x . Concept Used: Basic properties and rules for limits: for any real number c, suppose that functions f and g has limit at x = c , then Constant rule: lim x → c k = k for any constant k Sum rule : lim x → c ( f ( x ) + g ( x ) ) = lim x → c f ( x ) + lim x → c g ( x ) Product rule: lim x → c [ f ( x ) g ( x ) ] = ( lim x → c f ( x ) ) ( lim x → c g ( x ) ) Quotient rule : lim x → c [ f ( x ) g ( x ) ] = lim x → c f ( x ) lim x → c g ( x ) if lim x → c g ( x ) ≠ 0 If the limit if a function f ( x ) cannot be evaluated when x → c by direct substitution, try to reduce the function to another function that agrees with f ( x ) for all values of x except at the troublesome value x = c Special limit involving sine and cosine lim x → 0 sin x x = 1 , lim x → 0 cos x − 1 x = 0 Calculation: In order to evaluate the limit lim x → 0 cot 3 x cot x Since, by direct substitution of limit here we get lim x → 0 cot 3 x cot x = lim x → 0 cot 3 x lim x → 0 cot x using quoitent rule of limits = cot ( 0 ) cot 0 = ∞ ∞ And ∞ ∞ is an indeterminate form Thus, try to simplify the function and then evaluate the limit, as lim x → 0

7th Edition

ISBN: 9781524916817

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

Publisher: Kendall Hunt Publishing

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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 topic…

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Chapter 2.2, Problem 24PS

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To evaluate the given limit limx→0cot3xcotx .

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Chapter 2.2, Problem 23PS

Chapter 2.2, Problem 25PS

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Chapter 2 Solutions

Calculus

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To evaluate the given limit lim x → 0 cot 3 x cot x .

Solution Summary: The author explains the basic properties and rules for limits: for any real number c, suppose that functions f and g have limit at x=c.

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To evaluate the given limit limx→0cot3xcotx .

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Chapter 2 Solutions