Explanation Given information: f ( x , y ) = 2 x 2 + 3 y 2 x y Calculation: We will calculate limit of the function by setting y = m x as follows, Substituting y = m x in f ( x , y ) = 2 x 2 + 3 y 2 x y we get, f ( x , m x ) = 2 x 2 + 3 y 2 x y = 2 x 2 + 3 ( m x ) 2 x ( m x ) = 2 x 2 + 3 m 2 x 2 m x 2 = x 2 ( 2 + 3 m 2 ) m x 2 = 2 + 3 m 2 m Now we will find the limit o

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ISBN: 9781319050740

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

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Limits of Multivariable Functions

Multivariable calculus is an extension of single variable calculus. It involves several variables instead of just one. Assume there is an open set containing points (x0, y0), let f be a function defined in that open interval except for the points (x0, y0…

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Chapter 14.2, Problem 16E

To determine

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the limit of $f(x,y)$ depend on $m$ , and therefore the limit $lim(x,y)→(0,0)f(x,y)$ does not exist.

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Chapter 14.2, Problem 15E

Chapter 14.2, Problem 17E

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Chapter 14 Solutions

Calculus: Early Transcendentals

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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Toshow: the limit of f ( x , y ) depend on m , and therefore the limit lim ( x , y ) → ( 0 , 0 ) f ( x , y ) does not exist.

Solution Summary: The author explains that the limit of f(x,y) depend on m, and thus the function limit does not exist.

Concept explainers

Toshow:

the limit of $f(x,y)$ depend on $m$ , and therefore the limit $lim(x,y)→(0,0)f(x,y)$ does not exist.

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Chapter 14 Solutions