Step 1: Core Idea:
This question assesses limits and continuity of a two-variable function. The limit must be consistent along all paths to a point for it to exist. If the limit fails to exist, the function is neither continuous nor differentiable at that point.
Step 2: Method:
To check the limit at \((0,0)\), investigate the function's behavior along different paths. Polar coordinates, \( x = r\cos\theta, y = r\sin\theta \), are a useful tool. As \( (x,y) \to (0,0) \), then \( r \to 0 \).
Step 3: Step-by-step Solution:
Substitute polar coordinates into the function for \( (x,y) eq (0,0) \):
\[ f(r\cos\theta, r\sin\theta) = \frac{r\cos\theta}{\sqrt{(r\cos\theta)^2 + (r\sin\theta)^2}} = \frac{r\cos\theta}{\sqrt{r^2(\cos^2\theta + \sin^2\theta)}} = \frac{r\cos\theta}{\sqrt{r^2}} = \frac{r\cos\theta}{r} = \cos\theta \]Next, evaluate the limit as \( (x,y) \to (0,0) \), which translates to \( r \to 0 \):
\[ \lim_{r \to 0} f(r\cos\theta, r\sin\theta) = \lim_{r \to 0} \cos\theta = \cos\theta \]The limit's value depends on \( \theta \), the approach angle.
Examples:
- Along the positive x-axis (\( \theta=0 \)), the limit is \( \cos(0) = 1 \).
- Along the positive y-axis (\( \theta=\pi/2 \)), the limit is \( \cos(\pi/2) = 0 \).
Because the limit varies with the approach path, \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist.
Step 4: Conclusion:
- Since the limit does not exist at \((0,0)\), the function is not continuous there, eliminating options B and C.
- A removable discontinuity needs a limit that exists. Because the limit does not exist, the discontinuity is non-removable, eliminating option D.
- Therefore, only statement A is correct.
Final Answer: \( \lim_{(x,y) \to (0,0)} f(x,y) \) does not exist.