Question

The value of \(\lim_{(x,y) \to (0,0)} \frac{x^2y}{x^2 + y^2}\) is: 

• A. 1
• B. 0
• C. \(\infty\)
• D. Does not exist

Hint:

Exam Tip:
For limits of functions of two variables:

• Use polar coordinates if the expression involves \(x^2 + y^2\).
• The limit is 0 if the expression tends to 0 as \(r \to 0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to evaluate the limit of a function of two variables as \((x, y) \to (0, 0)\).

Step 2: Key Formula or Approach:
We can use polar coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\).
Then \(x^2 + y^2 = r^2\), and \(x^2 y = r^2 \cos^2 \theta \cdot r \sin \theta = r^3 \cos^2 \theta \sin \theta\).

Step 3: Detailed Explanation:
\[ \frac{x^2 y}{x^2 + y^2} = \frac{r^3 \cos^2 \theta \sin \theta}{r^2} = r \cos^2 \theta \sin \theta \] As \((x, y) \to (0, 0)\), \(r \to 0\).
Since \(|\cos^2 \theta \sin \theta| \le 1\), the limit is 0.
So, the limit exists and is equal to 0.

Step 4: Final Answer:
Therefore, option (B) is correct.