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Q27E

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Found in: Page 632

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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.

See the step by step solution

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.