Question

Let \( f: \mathbb{R} \times \mathbb{R} \to \mathbb{R} \) be defined as:

\[ f(x, y) = \begin{cases} \frac{x}{\sqrt{x^2 + y^2}} & \text{if } (x, y) \neq (0, 0) \\ 1 & \text{if } (x, y) = (0, 0) \end{cases} \] Then, which of the following statements is true?

• A. \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist
• B. \( f(x,y) \) is continuous but not differentiable
• C. \( f(x,y) \) is differentiable function
• D. \( f(x,y) \) have removable discontinuity

Hint:

When checking the limit of a multivariable function at the origin, converting to polar coordinates is a very effective strategy. If the resulting expression depends on \( \theta \) after \( r \) has been taken to 0, the limit does not exist.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question tests the concepts of limits and continuity for a function of two variables. For the limit to exist at a point, the function must approach the same value along all possible paths to that point. If the limit does not exist, the function cannot be continuous or differentiable at that point.

Step 2: Key Formula or Approach:
To check for the existence of the limit at \((0,0)\), we can test the value of the function as we approach the origin along different paths. A common technique is to use the path \(y=mx\) or to switch to polar coordinates. Let's use polar coordinates: \( x = r\cos\theta, y = r\sin\theta \). As \( (x,y) \to (0,0) \), we have \( r \to 0 \).

Step 3: Detailed Explanation:
Substitute the polar coordinates into the function for \( (x,y) \neq (0,0) \): \[ f(r\cos\theta, r\sin\theta) = \frac{r\cos\theta}{\sqrt{(r\cos\theta)^2 + (r\sin\theta)^2}} = \frac{r\cos\theta}{\sqrt{r^2(\cos^2\theta + \sin^2\theta)}} = \frac{r\cos\theta}{\sqrt{r^2}} = \frac{r\cos\theta}{r} = \cos\theta \] Now, let's evaluate the limit as \( (x,y) \to (0,0) \), which corresponds to \( r \to 0 \): \[ \lim_{r \to 0} f(r\cos\theta, r\sin\theta) = \lim_{r \to 0} \cos\theta = \cos\theta \] The result of the limit depends on \( \theta \), which represents the angle of approach to the origin. For example: - If we approach along the positive x-axis (\( \theta=0 \)), the limit is \( \cos(0) = 1 \). - If we approach along the positive y-axis (\( \theta=\pi/2 \)), the limit is \( \cos(\pi/2) = 0 \). Since the limit has different values for different paths of approach, the overall limit \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist.

Step 4: Analyzing the Consequences:
- Since the limit does not exist at \((0,0)\), the function is not continuous at \((0,0)\). This rules out options B and C. - A removable discontinuity occurs when the limit exists but is not equal to the function's value. Since the limit does not exist, the discontinuity is non-removable. This rules out option D. - Therefore, statement A is the only correct one.

Final Answer: \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist.