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Chapter 12: Problem 29

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

Short Answer

Expert verified
Question: Prove that the limit $$\lim_{(x, y) \rightarrow (0,0)} \frac{y^{4}-2 x^{2}}{y^{4}+x^{2}}$$ does not exist using the Two-Path Test. Answer: The limit does not exist because the limit along the y = 0 path is -2 and the limit along the x = 0 path is 1. Since the limits along these two different paths are not equal, the limit does not exist, as shown by the Two-Path Test.

Step by step solution

01

Identify two different paths

For this problem, we will approach the limit point (0,0) along the following paths: 1. y = 0 2. x = 0
02

Compute the limit along the y = 0 path

First, we will take the limit as (x, y) 鈫� (0, 0) along the y = 0 path. Substitute y = 0 into the given expression and simplify: $$\lim _{(x, 0) \rightarrow(0,0)} \frac{0^{4}-2 x^{2}}{0^{4}+x^{2}} = \lim_{x \rightarrow 0} \frac{-2x^2}{x^2}$$ Since the numerator and denominator share a common term of x虏, cancel out the term: $$\lim_{x \rightarrow 0} \frac{-2x^2}{x^2} = \lim_{x \rightarrow 0} -2$$ The limit is -2 along the y = 0 path.
03

Compute the limit along the x = 0 path

Next, we will take the limit as (x, y) 鈫� (0, 0) along the x = 0 path. Substitute x = 0 into the given expression and simplify: $$\lim _{(0, y) \rightarrow(0,0)} \frac{y^{4}-2 \cdot 0^{2}}{y^{4}+0^{2}} = \lim_{y \rightarrow 0} \frac{y^4}{y^4}$$ Since the numerator and denominator share a common term of y鈦�, cancel out the term: $$\lim_{y \rightarrow 0} \frac{y^4}{y^4} = \lim_{y \rightarrow 0} 1$$ The limit is 1 along the x = 0 path.
04

Conclude using the Two-Path Test

Since we have found that the limit along the y = 0 path (-2) is different from the limit along the x = 0 path (1), we can conclude, by the Two-Path Test, that the given limit does not exist: $$\lim _{(x, y) \rightarrow(0,0)} \frac{y^{4}-2 x^{2}}{y^{4}+x^{2}}$$ does not exist.

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