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

Loose-leaf Version For Calculus: Early Transcendentals

4th Edition

ISBN: 9781319055912

Author: Jon Rogawski, Colin Adams, Robert Franzosa

Publisher: W. H. Freeman

See similar textbooks

Concept explainers

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…

Question

Chapter 14.2, Problem 16E

To determine

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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.2, Problem 15E

Chapter 14.2, Problem 17E

Students have asked these similar questions

A function is defined on the interval (-π/2,π/2) by this multipart rule: if -π/2 < x < 0 f(x) = a if x=0 31-tan x +31-cot x if 0 < x < π/2 Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0. a= b= 3

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a. f(x) = (x + 4x4) 5, a = -1 lim f(x) X--1 = lim x+4x X--1 lim X-1 4 x+4x 5 ))" 5 )) by the power law by the sum law lim (x) + lim X--1 4 4x X-1 -(0,00+( Find f(-1). f(-1)=243 lim (x) + -1 +4 35 4 ([ ) lim (x4) 5 x-1 Thus, by the definition of continuity, f is continuous at a = -1. by the multiple constant law by the direct substitution property

1. Compute Lo F⚫dr, where and C is defined by F(x, y) = (x² + y)i + (y − x)j r(t) = (12t)i + (1 − 4t + 4t²)j from the point (1, 1) to the origin.

Chapter 14 Solutions

Loose-leaf Version For 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

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

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.

Want to see the full answer?

Chapter 14 Solutions

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.

Want to see the full answer?

Chapter 14 Solutions

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.