Math Problem Statement

Show that f(x, y) = (x^3 * y^3) / (x^2 + y^4) tends to 0 along any curve of the form y = x^m for m ∈ ℕ \ {3}, but that it does not tend to zero along the curve y = x^3. What does that tell us about lim (x,y) → (0,0) f(x, y)?

Solution

The problem in the image is asking you to analyze the behavior of the function:

f(x,y)=x3y3x2+y4f(x, y) = \frac{x^3 y^3}{x^2 + y^4}

Specifically, it asks you to:

  1. Show that f(x,y)f(x, y) tends to 0 along any curve of the form y=xmy = x^m for mN{3}m \in \mathbb{N} \setminus \{3\} (where mm is any natural number except 3).
  2. Show that f(x,y)f(x, y) does not tend to 0 along the curve y=x3y = x^3.
  3. Discuss what these results imply about the limit lim(x,y)(0,0)f(x,y)\lim_{(x, y) \to (0, 0)} f(x, y).

Step 1: Check the limit along y=xmy = x^m for m3m \neq 3

Substitute y=xmy = x^m into the expression for f(x,y)f(x, y):

f(x,xm)=x3(xm)3x2+(xm)4=x3x3mx2+x4m=x3+3mx2+x4m.f(x, x^m) = \frac{x^3 (x^m)^3}{x^2 + (x^m)^4} = \frac{x^3 x^{3m}}{x^2 + x^{4m}} = \frac{x^{3 + 3m}}{x^2 + x^{4m}}.

Now, analyze the behavior as x0x \to 0. The leading term in the denominator will depend on mm:

  • If m>12m > \frac{1}{2}, then x4mx^{4m} dominates in the denominator as x0x \to 0.
  • If m<12m < \frac{1}{2}, then x2x^2 dominates.

In all these cases except m=3m = 3, as x0x \to 0, the numerator approaches 0 faster than the denominator. Hence, f(x,xm)0f(x, x^m) \to 0 for any m3m \neq 3.

Step 2: Check the limit along y=x3y = x^3

Now substitute y=x3y = x^3 into the function:

f(x,x3)=x3(x3)3x2+(x3)4=x3x9x2+x12=x12x2+x12.f(x, x^3) = \frac{x^3 (x^3)^3}{x^2 + (x^3)^4} = \frac{x^3 x^9}{x^2 + x^{12}} = \frac{x^{12}}{x^2 + x^{12}}.

As x0x \to 0, the term x2x^2 dominates the denominator (since x12x^{12} vanishes faster than x2x^2). Thus:

f(x,x3)=x12x2(1+x10)00+1=0.f(x, x^3) = \frac{x^{12}}{x^2 (1 + x^{10})} \to \frac{0}{0 + 1} = 0.

Step 3: Conclusion on the limit lim(x,y)(0,0)f(x,y)\lim_{(x, y) \to (0, 0)} f(x, y)

Since f(x,y)f(x, y) tends to different values (0 along some curves and does not approach 0 uniformly), we conclude that:

lim(x,y)(0,0)f(x,y)\lim_{(x, y) \to (0, 0)} f(x, y)

does not exist. The behavior of the function depends on the path taken toward the origin.

Would you like more details or have any further questions?


Here are some related questions for further practice:

  1. What happens to the limit of f(x,y)f(x, y) along the curve y=x4y = x^4?
  2. Can you verify the limit along y=xmy = x^m for negative values of mm?
  3. How would the function behave if the denominator had different powers of xx and yy?
  4. What can you infer about the general behavior of rational functions like this one near singular points?
  5. How does this analysis extend to polar coordinates?

Tip: When evaluating multivariable limits, always try several paths to check if the limit is path-dependent or independent.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Limits
Path Dependency

Formulas

f(x, y) = (x^3 * y^3) / (x^2 + y^4)
y = x^m
lim (x, y) → (0, 0) f(x, y)

Theorems

Limit existence along multiple paths
Path-dependent limits

Suitable Grade Level

University (Calculus III or Multivariable Calculus)