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Chapter 13: 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
Short answer: Since the limits along the paths y=0 and x=0 are not the same (-2 and 1, respectively), by the Two-Path Test, the limit does not exist when (x, y) approaches (0,0).

Step by step solution

01

Approach along the path y=0

Along the path y=0, we substitute y=0 into the function: $$\frac{y^{4}-2x^{2}}{y^{4}+x^{2}} = \frac{(0)^{4}-2x^{2}}{(0)^{4}+x^{2}} = \frac{-2x^{2}}{x^{2}} = -2$$ In this case, the limit as (x, y) approaches (0,0) is -2.
02

Approach along the path x=0

Along the path x=0, we substitute x=0 into the function: $$\frac{y^{4}-2x^{2}}{y^{4}+x^{2}} = \frac{y^{4}-2(0)^{2}}{y^{4}+(0)^{2}} = \frac{y^{4}}{y^{4}} = 1$$ In this case, the limit as (x, y) approaches (0,0) is 1.
03

Compare the limits along different paths

Since the limit is -2 along the path y=0 and 1 along the path x=0, the two limits are not the same. Thus, by the Two-Path Test, the limit does not exist when (x,y) approaches (0,0). Therefore, we conclude that: $$\lim _{(x, y) \rightarrow(0,0)} \frac{y^{4}-2 x^{2}}{y^{4}+x^{2}} \text{ does not exist.}$$

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Most popular questions from this chapter

Use the formal definition of a limit to prove that $$\lim _{(x, y) \rightarrow(a, b)}(f(x, y)+g(x, y))=\lim _{(x, y) \rightarrow(a, b)} f(x, y)+$$ $$\lim _{(x, y) \rightarrow(a, b)} g(x, y)$$

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2}}{x^{2}+y^{2}}$$

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