Find two paths of approach from which one can conclude that the function has no limit as (x, y) approaches (0, 0).
-$\text { Define } f ( 0,0 ) \text { in a way that extends } f ( x , y ) = \frac { 6 x ^ { 2 } - x ^ { 2 } y + 6 y ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } \text { to be continuous at the origin. }$
A) f(0, 0) = 0
B) f(0, 0) = 6
C) f(0, 0) = 2
D) f(0, 0) = 12

Question

Quizplus · Accepted Answer

CTo determine if a function has a limit as $(x, y) \rightarrow (0, 0)$, we can approach the origin along different paths. If the limit depends on the path, then the function does not have a limit at that point.For the given function $f(x, y) = \frac{6x^2 - x^2y + 6y^2}{x^2 + y^2}$, let's consider two paths:1. Along the x-axis, where $y = 0$. Substituting $y = 0$ into the function gives $f(x, 0) = \frac{6x^2}{x^2} = 6$, suggesting that if $f(0, 0)$ were to make $f$ continuous at the origin, it should be defined as $f(0, 0) = 6$.2. Along the line $y = x$. Substituting $y = x$ into the function gives $f(x, x) = \frac{6x^2 - x^3 + 6x^2}{x^2 + x^2} = \frac{12x^2 - x^3}{2x^2} = 6 - \frac{1}{2}x$. As $(x, x) \rightarrow (0, 0)$, this expression approaches $6$, consistent with the path along the x-axis.However, to demonstrate that the function does not have a limit as $(x, y) \rightarrow (0, 0)$, we need to find paths that yield different limits. A mistake was made in suggesting two paths that both approach the same value. Instead, consider:- Along the y-axis, where $x = 0$. Substituting $x = 0$ into the function gives $f(0, y) = \frac{6y^2}{y^2} = 6$, which is consistent with the previous path.To correctly identify paths that show the function does not have a limit, we should consider paths such as $y = mx$ (where $m$ is a constant) or using polar coordinates to show how the limit can vary with the angle of approach. However, the initial explanation incorrectly suggests that the function does not have a limit based on the paths provided. The correct approach to extend $f(x, y)$ to be continuous at the origin would be to define $f(0, 0) = 6$, as this is the value the function approaches from any radial direction from the origin, not demonstrating the function's discontinuity or the existence of multiple limits. The error in the explanation and the paths chosen does not correctly justify the selection of options B and C. The correct approach should involve demonstrating that the limit is the same from any direction, which would support defining $f(0, 0) = 6$ for continuity, not to show discontinuity or multiple limits.

Find Two Paths of Approach from Which One Can Conclude