Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Determine the set of points at which the function is continuous: \(f(x,y) = \left\{ \begin{aligned}{l}\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}, if(x,y) \ne (0,0)\\1, if(x,y) = (0,0)\end{aligned} \right.\)

\(\left\{ {(x,y) \in {R^2}|(x,y) \ne (0,0)} \right\}\) is the set of points at which the function is continuous.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Finding limit of function along x-axis (\(y = 0\)):

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}{{(0)}^3}}}{{2{x^2} + {{(0)}^2}}}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{0}{{2{x^2} + 0}}} \right) = 0\)

Thus, the limit of function along x-axis is zero.

Step 2: Finding limit of function along y-axis

The limit of function along y-axis (\(x = 0\)) is

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}} \right)\)

\( = \mathop {\lim }\limits_{y \to 0} \left( {\frac{{{{(0)}^2}{y^3}}}{{2(0) + {y^2}}}} \right) = 0\)

Thus, the limit of function along y-axis is zero.

Step 3: Finding limit of function along the line y-axis:

\(\mathop {\lim }\limits_{(x,y) \to (0,0)} \left( {\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{x^2}{{(mx)}^3}}}{{2{x^2} + {{(mx)}^2}}}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{m^3}{x^5}}}{{2{x^2} + {m^2}{x^2}}}} \right)\)

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{m^3}{x^5}}}{{{x^2}(2 + {m^2})}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{m^3}{x^3}}}{{2 + {m^2}}}} \right) = 0\)

Thus, for any value of \(m\) the limit always approximate to 0. Hence, the limit of the function at \((0,0)\) is o because along different path, the function value always approaches to zero.

Step 4: Removing the discontinuity:

Also, the function value \((0,0)\)is 1. That is \(f(0,0) = 1\)by the definition of the given function. Thus, \(\mathop {\lim }\limits_{(x,y) \to (0,0)} f(x,y) \ne (0,0)\)implies that function is not continuous at \((0,0)\).

Further, to remove the point of discontinuity from the \(xy\)-plane in order to determine the set of points at which the function is continuous. As \((0,0)\)is the only point of discontinuity in \(xy\)-plane.

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Essential Calculus: Early Transcendentals

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Short Answer

Determine the set of points at which the function is continuous: \(f(x,y) = \left\{ \begin{aligned}{l}\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}, if(x,y) \ne (0,0)\\1, if(x,y) = (0,0)\end{aligned} \right.\)

Step by Step Solution

Step 1: Finding limit of function along x-axis (\(y = 0\)):

Step 2: Finding limit of function along y-axis

Step 3: Finding limit of function along the line y-axis:

Step 4: Removing the discontinuity:

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Determine the set of points at which the function is continuous: \(f(x,y) = \left\{ \begin{aligned}{l}\frac{{{x^2}{y^3}}}{{2{x^2} + {y^2}}}, if(x,y) \ne (0,0)\\1, if(x,y) = (0,0)\end{aligned} \right.\)

Step by Step Solution

Step 1: Finding limit of function along x-axis (\(y = 0\)):

Step 2: Finding limit of function along y-axis

Step 3: Finding limit of function along the line y-axis:

Step 4: Removing the discontinuity:

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