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Chapter 12: Problem 28

Use the Two-Path Test to prove that the following limits do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{4 x y}{3 x^{2}+y^{2}}$$

Short Answer

Expert verified
Answer: No, the limit does not exist as the point (0,0) is approached. We found different limits along different paths, indicating that the limit does not exist at this point.

Step by step solution

01

Path 1: Approach along the x-axis

Along the x-axis, y = 0. We can substitute this into our function and simplify: $$\lim_{(x, y) \rightarrow (0,0)} \frac{4xy}{3x^2 + y^2} = \lim_{x \rightarrow 0} \frac{4x(0)}{3x^2 + (0)^2} = \lim_{x \rightarrow 0} \frac{0}{3x^2} = 0$$
02

Path 2: Approach along the y-axis

Along the y-axis, x = 0. We can substitute this into our function and simplify: $$\lim_{(x, y) \rightarrow (0,0)} \frac{4xy}{3x^2 + y^2} = \lim_{y \rightarrow 0} \frac{4(0)y}{3(0)^2 + y^2} = \lim_{y \rightarrow 0} \frac{0}{y^2} = 0$$
03

Path 3: Approach along the line y = x

Along the line y = x, we can substitute y = x into our function and simplify: $$\lim_{(x, y) \rightarrow (0,0)} \frac{4xy}{3x^2 + y^2} = \lim_{x \rightarrow 0} \frac{4x(x)}{3x^2 + x^2} = \lim_{x \rightarrow 0} \frac{4x^2}{4x^2} = 1$$ Since the limit along Path 1 and Path 2 is 0, while the limit along Path 3 is 1, the limits along these paths are not equal and thus the limit does not exist.

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