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Sagot :
To determine the set of points at which the function
[tex]\[ f(x, y) = \begin{cases} \frac{x y}{x^2 + x y + y^2} & \text{if } (x, y) \ne (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases} \][/tex]
is continuous, we need to find where this function does not break continuity.
1. Analyzing Continuity at Non-Origin Points [tex]\((x, y)\)[/tex]:
First, consider the function for points [tex]\((x, y) \ne (0, 0)\)[/tex].
[tex]\[ f(x, y) = \frac{x y}{x^2 + x y + y^2} \][/tex]
For [tex]\((x, y) \ne (0, 0)\)[/tex], the function is defined by a ratio of polynomials, which are continuous everywhere in their domain, provided the denominator does not equal zero.
2. Denominator Analysis:
Note that [tex]\(x^2 + x y + y^2\)[/tex] is always non-zero for [tex]\((x, y) \ne (0, 0)\)[/tex]. This is because:
- If [tex]\(x = 0\)[/tex] and [tex]\(y \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2 = y^2 \ne 0\)[/tex].
- If [tex]\(y = 0\)[/tex] and [tex]\(x \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2 = x^2 \ne 0\)[/tex].
- If [tex]\(x \ne 0\)[/tex] and [tex]\(y \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2\)[/tex] is a sum of positive terms and cannot be zero.
3. Continuity at the Origin [tex]\((0, 0)\)[/tex]:
To check for the continuity at [tex]\((0, 0)\)[/tex], we need to see if the limit as [tex]\((x, y)\)[/tex] approaches [tex]\((0, 0)\)[/tex] equals the function's value at [tex]\((0, 0)\)[/tex]. Let's examine the limit:
[tex]\[ \lim_{(x, y) \to (0, 0)} \frac{x y}{x^2 + x y + y^2} \][/tex]
By trying different paths:
- Along [tex]\(y = kx\)[/tex], the expression simplifies to:
[tex]\[ \frac{x(kx)}{x^2 + x(kx) + (kx)^2} = \frac{kx^2}{x^2 + kx^2 + k^2x^2} = \frac{kx^2}{(1+k+k^2)x^2} = \frac{k}{1+k+k^2} \][/tex]
This limit depends on [tex]\(k\)[/tex], hence the limit is not unique as [tex]\((x, y) \to (0, 0)\)[/tex].
Since the limit is not unique, we conclude that:
[tex]\[ \lim_{(x, y) \to (0, 0)} \frac{x y}{x^2 + x y + y^2} \ne 0 \][/tex]
Therefore, the function is discontinuous at [tex]\((0, 0)\)[/tex].
Given these observations, the function [tex]\( f(x, y) \)[/tex] is continuous everywhere except at [tex]\((0, 0)\)[/tex].
Therefore, the set of points where the function is continuous is:
[tex]\[ \{(x, y) \mid (x, y) \ne (0,0)\} \][/tex]
[tex]\[ f(x, y) = \begin{cases} \frac{x y}{x^2 + x y + y^2} & \text{if } (x, y) \ne (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases} \][/tex]
is continuous, we need to find where this function does not break continuity.
1. Analyzing Continuity at Non-Origin Points [tex]\((x, y)\)[/tex]:
First, consider the function for points [tex]\((x, y) \ne (0, 0)\)[/tex].
[tex]\[ f(x, y) = \frac{x y}{x^2 + x y + y^2} \][/tex]
For [tex]\((x, y) \ne (0, 0)\)[/tex], the function is defined by a ratio of polynomials, which are continuous everywhere in their domain, provided the denominator does not equal zero.
2. Denominator Analysis:
Note that [tex]\(x^2 + x y + y^2\)[/tex] is always non-zero for [tex]\((x, y) \ne (0, 0)\)[/tex]. This is because:
- If [tex]\(x = 0\)[/tex] and [tex]\(y \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2 = y^2 \ne 0\)[/tex].
- If [tex]\(y = 0\)[/tex] and [tex]\(x \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2 = x^2 \ne 0\)[/tex].
- If [tex]\(x \ne 0\)[/tex] and [tex]\(y \ne 0\)[/tex], then [tex]\(x^2 + x y + y^2\)[/tex] is a sum of positive terms and cannot be zero.
3. Continuity at the Origin [tex]\((0, 0)\)[/tex]:
To check for the continuity at [tex]\((0, 0)\)[/tex], we need to see if the limit as [tex]\((x, y)\)[/tex] approaches [tex]\((0, 0)\)[/tex] equals the function's value at [tex]\((0, 0)\)[/tex]. Let's examine the limit:
[tex]\[ \lim_{(x, y) \to (0, 0)} \frac{x y}{x^2 + x y + y^2} \][/tex]
By trying different paths:
- Along [tex]\(y = kx\)[/tex], the expression simplifies to:
[tex]\[ \frac{x(kx)}{x^2 + x(kx) + (kx)^2} = \frac{kx^2}{x^2 + kx^2 + k^2x^2} = \frac{kx^2}{(1+k+k^2)x^2} = \frac{k}{1+k+k^2} \][/tex]
This limit depends on [tex]\(k\)[/tex], hence the limit is not unique as [tex]\((x, y) \to (0, 0)\)[/tex].
Since the limit is not unique, we conclude that:
[tex]\[ \lim_{(x, y) \to (0, 0)} \frac{x y}{x^2 + x y + y^2} \ne 0 \][/tex]
Therefore, the function is discontinuous at [tex]\((0, 0)\)[/tex].
Given these observations, the function [tex]\( f(x, y) \)[/tex] is continuous everywhere except at [tex]\((0, 0)\)[/tex].
Therefore, the set of points where the function is continuous is:
[tex]\[ \{(x, y) \mid (x, y) \ne (0,0)\} \][/tex]
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