Math Problem Statement

Show that lim (x,y)➡️(0,0) xy/x²+y² do not exist

Solution

To prove that the limit does not exist for the function

f(x,y)=xyx2+y2f(x, y) = \frac{xy}{x^2 + y^2}

as (x,y)(0,0)(x, y) \to (0, 0), we can attempt to approach (0,0)(0, 0) along different paths and see if the limit depends on the path chosen. If the limit varies based on the path, it means that the overall limit does not exist.

Approach 1: Along the x-axis (y=0y = 0)

Substitute y=0y = 0 into the function:

f(x,0)=x0x2+02=0f(x, 0) = \frac{x \cdot 0}{x^2 + 0^2} = 0

So, along the x-axis, the function tends to 0 as x0x \to 0.

Approach 2: Along the y-axis (x=0x = 0)

Substitute x=0x = 0 into the function:

f(0,y)=0y02+y2=0f(0, y) = \frac{0 \cdot y}{0^2 + y^2} = 0

Along the y-axis, the function also tends to 0 as y0y \to 0.

Approach 3: Along the line y=xy = x

Substitute y=xy = x into the function:

f(x,x)=xxx2+x2=x22x2=12f(x, x) = \frac{x \cdot x}{x^2 + x^2} = \frac{x^2}{2x^2} = \frac{1}{2}

Along the line y=xy = x, the function tends to 12\frac{1}{2} as x0x \to 0.

Approach 4: Along the line y=xy = -x

Substitute y=xy = -x into the function:

f(x,x)=x(x)x2+(x)2=x22x2=12f(x, -x) = \frac{x \cdot (-x)}{x^2 + (-x)^2} = \frac{-x^2}{2x^2} = \frac{-1}{2}

Along the line y=xy = -x, the function tends to 12-\frac{1}{2} as x0x \to 0.

Conclusion

We have found different limits along different paths:

  • Along the x-axis, the limit is 0.
  • Along the y-axis, the limit is 0.
  • Along the line y=xy = x, the limit is 12\frac{1}{2}.
  • Along the line y=xy = -x, the limit is 12-\frac{1}{2}.

Since the limit depends on the path chosen, the overall limit does not exist.

Would you like more details on any part of this explanation?

Here are 5 related questions to explore this further:

  1. What is the formal definition of a two-variable limit?
  2. How do you prove the existence of a multivariable limit?
  3. What happens if you approach the origin in polar coordinates?
  4. Can limits along curves other than straight lines change the result?
  5. What is the concept of continuity in two-variable functions?

Tip: In multivariable calculus, limits depend not only on the point but also on the path of approach. Always test multiple paths when trying to prove or disprove the existence of a limit.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Limits of Multivariable Functions
Path Dependence in Limits

Formulas

f(x, y) = (xy) / (x² + y²)

Theorems

Path Dependence Theorem in Multivariable Limits

Suitable Grade Level

Undergraduate Calculus (College Level)