Math Problem Statement

Evaluate the limit \( \lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4} \)

Solution

To evaluate the limit

lim(x,y)(0,0)y2sin2xx4+y4\lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4}

we'll analyze how the expression behaves as (x,y)(x, y) approaches (0,0)(0, 0) from different paths. Often, using polar coordinates or testing specific paths can help reveal if the limit exists.

Step-by-Step Solution

  1. Convert to Polar Coordinates: In polar coordinates, x=rcosθx = r \cos \theta and y=rsinθy = r \sin \theta. Then, x2+y2=r2x^2 + y^2 = r^2, and the function transforms accordingly: y2=r2sin2θandx4+y4=r4(cos4θ+sin4θ).y^2 = r^2 \sin^2 \theta \quad \text{and} \quad x^4 + y^4 = r^4 (\cos^4 \theta + \sin^4 \theta).

    Substitute into the expression: y2sin2xx4+y4=r2sin2θsin2(rcosθ)r4(cos4θ+sin4θ).\frac{y^2 \sin^2 x}{x^4 + y^4} = \frac{r^2 \sin^2 \theta \sin^2 (r \cos \theta)}{r^4 (\cos^4 \theta + \sin^4 \theta)}.

  2. Simplify the Expression: Simplify by canceling r2r^2 from the numerator and denominator: =sin2θsin2(rcosθ)r2(cos4θ+sin4θ).= \frac{\sin^2 \theta \sin^2 (r \cos \theta)}{r^2 (\cos^4 \theta + \sin^4 \theta)}.

  3. Analyze the Behavior of sin2(rcosθ)\sin^2 (r \cos \theta) as r0r \to 0: Notice that sin2(rcosθ)\sin^2 (r \cos \theta) behaves approximately like (rcosθ)2(r \cos \theta)^2 for small rr. Thus, sin2(rcosθ)r2cos2θ.\sin^2 (r \cos \theta) \approx r^2 \cos^2 \theta.

    Substitute this approximation: sin2θr2cos2θr2(cos4θ+sin4θ)=r2sin2θcos2θr2(cos4θ+sin4θ).\frac{\sin^2 \theta \cdot r^2 \cos^2 \theta}{r^2 (\cos^4 \theta + \sin^4 \theta)} = \frac{r^2 \sin^2 \theta \cos^2 \theta}{r^2 (\cos^4 \theta + \sin^4 \theta)}.

    Cancel r2r^2: =sin2θcos2θcos4θ+sin4θ.= \frac{\sin^2 \theta \cos^2 \theta}{\cos^4 \theta + \sin^4 \theta}.

  4. Check if the Result Depends on θ\theta: This expression depends on θ\theta because sin2θcos2θ\sin^2 \theta \cos^2 \theta and cos4θ+sin4θ\cos^4 \theta + \sin^4 \theta vary with θ\theta. Therefore, the limit depends on the path taken to approach (0,0)(0,0), which means the limit does not exist.

Conclusion

Since the limit depends on the angle θ\theta, it does not converge to a single value as (x,y)(0,0)(x, y) \to (0,0). Thus:

lim(x,y)(0,0)y2sin2xx4+y4does not exist.\lim_{(x, y) \to (0, 0)} \frac{y^2 \sin^2 x}{x^4 + y^4} \quad \text{does not exist}.

Would you like further details or have questions about any specific steps?


Here are some related questions to further explore this topic:

  1. What is the importance of converting to polar coordinates in multivariable limits?
  2. How can testing different paths help in determining the existence of a limit?
  3. Can we use other coordinate transformations, like spherical coordinates, for limits in higher dimensions?
  4. What are some common cases where limits do not exist due to path dependence?
  5. How does L'Hospital's Rule extend (if at all) to multivariable limits?

Tip: When checking multivariable limits, try substituting polar coordinates or testing specific paths. It’s often helpful in determining if a limit exists or depends on the approach path.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Limits
Polar Coordinates
Path Dependence

Formulas

Conversion to polar coordinates: \( x = r \cos \theta \) and \( y = r \sin \theta \)
Trigonometric approximation: \( \sin^2 (r \cos \theta) \approx (r \cos \theta)^2 \)

Theorems

Limit does not exist due to path dependence

Suitable Grade Level

Undergraduate Calculus