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Chapter 12: Problem 30

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

Short Answer

Expert verified
Question: Use the Two-Path Test to prove that the following limit does not exist: $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}-y^{2}}{x^{3}+y^{2}}$$ Answer: The limit does not exist. Explanation: By approaching the limit along the x-axis, we found that the limit is 1, and by approaching the limit along the line y = x, we found that the limit is -1. Since the limits are different along the two paths, according to the Two-Path Test, the given limit does not exist.

Step by step solution

01

Approach the limit along the x-axis

In order to find the limit along the x-axis, we will set y = 0. This will simplify the function as follows: $$\lim_{(x,0) \rightarrow (0,0)} \frac{x^{3}-0^{2}}{x^{3}+0^{2}} = \lim_{x \rightarrow 0} \frac{x^{3}}{x^{3}}$$ The function now simplifies to 1, so the limit along the x-axis is: $$\lim_{x \rightarrow 0} 1 = 1$$
02

Approach the limit along the line y = x

Now, we will find the limit along the line y = x. We do this by setting y = x and substituting it into the function: $$\lim_{(x,x) \rightarrow(0,0)} \frac{x^{3}-x^{2}}{x^{3}+x^{2}} $$ Now, we can factor out x虏 from both the numerator and the denominator: $$\lim_{x \rightarrow 0} \frac{x^{2}(x-1)}{x^{2}(x+1)} = \lim _{x \rightarrow 0} \frac{x-1}{x+1}$$ To find the limit, substitute x = 0 and simplify: $$\lim_{x \rightarrow 0} \frac{0-1}{0+1} = -1$$
03

Compare the limits from the two paths

From the previous steps, we found that the limit along the x-axis is 1 and the limit along the line y = x is -1. Since these two limits are different, we can conclude that the limit does not exist: $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}-y^{2}}{x^{3}+y^{2}}$$ does not exist according to the Two-Path Test.

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