We want to show that the limit $\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2}$ does not exist. We will do this by approaching $(0,0)$ along two different paths and showing that the limit is different along these paths. The two paths are the x-axis ($y=0$) and the line $y=x$.

AnalysisLimitsMultivariable CalculusPathwise LimitsContinuity
2025/4/28

1. Problem Description

We want to show that the limit lim(x,y)(0,0)xyx2+y2\lim_{(x,y)\to(0,0)} \frac{xy}{x^2+y^2} does not exist. We will do this by approaching (0,0)(0,0) along two different paths and showing that the limit is different along these paths. The two paths are the x-axis (y=0y=0) and the line y=xy=x.

2. Solution Steps

First, let's consider the path along the x-axis, where y=0y=0. Then the expression becomes:
limx0x(0)x2+(0)2=limx00x2=limx00=0\lim_{x\to 0} \frac{x(0)}{x^2 + (0)^2} = \lim_{x\to 0} \frac{0}{x^2} = \lim_{x\to 0} 0 = 0.
So along the x-axis, the limit is
0.
Next, let's consider the path along the line y=xy=x. Then the expression becomes:
limx0x(x)x2+x2=limx0x22x2=limx012=12\lim_{x\to 0} \frac{x(x)}{x^2 + x^2} = \lim_{x\to 0} \frac{x^2}{2x^2} = \lim_{x\to 0} \frac{1}{2} = \frac{1}{2}.
So along the line y=xy=x, the limit is 12\frac{1}{2}.
Since the limit is 0 along the x-axis and 12\frac{1}{2} along the line y=xy=x, the limit does not exist.

3. Final Answer

The limit does not exist.

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