Problem 28 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Show using the Two-Path Test that the limit $\lim _{(x, y) \rightarrow(0,0)} \frac{4 x y}{3 x^{2}+y^{2}}$ does not exist. Answer: The limit does not exist since along the path $y=x$, the limit is 1, and along the path $x=0$, the limit is 0. These different limits indicate that the overall limit does not exist.

Step 1: Write down the function

The given function is $$f(x, y) = \frac{4xy}{3x^2 + y^2}$$ and we want to compute the limit of this function as \((x, y) \rightarrow (0, 0)\).

Step 2: Choose the first path

Let's start by considering a path to \((0,0)\) along the line \(y = x\). Substitute \(y = x\) in the function: $$f(x, x) = \frac{4x^2}{3x^2 + x^2} = \frac{4x^2}{4x^2}$$

Step 3: Compute the limit along the first path

As \(x\) approaches \(0\), the value of \(\frac{4x^2}{4x^2}\) becomes: $$\lim_{x \rightarrow 0} \frac{4x^2}{4x^2} = 1$$

Step 4: Choose the second path

Now let's consider a path to \((0,0)\) along the line \(x = 0\). Substitute \(x = 0\) in the function: $$f(0, y) = \frac{4(0)y}{3(0)^2 + y^2} = 0$$

Step 5: Compute the limit along the second path

As \(y\) approaches \(0\), the value of \(0\) stays the same: $$\lim_{y \rightarrow 0} 0 = 0$$

Step 6: Compare the limits

We have found that the limit of the function is different when taking different paths to \((0, 0)\): $$\lim_{x \rightarrow 0} f(x, x) = 1 \neq 0 = \lim_{y \rightarrow 0} f(0, y)$$

Step 7: Conclusion

Since the limit of the function is different when taking different paths to \((0, 0)\), we can conclude using the Two-Path Test that the limit does not exist: $$\lim _{(x, y) \rightarrow(0,0)} \frac{4 x y}{3 x^{2}+y^{2}} \text{ does not exist}$$