Math Problem Statement
Examine the continuity of f : R2 → R at (0, 0), where for all (x, y) ∈ R2,f(x, y) :=
sin x sin
1
y
!
if y ̸= (0, 0),
0 if y = 0.
Solution
We are asked to examine the continuity of the function at , where is defined as follows:
\begin{cases} \sin(x) \sin\left(\frac{1}{y}\right), & \text{if } y \neq 0, \\ 0, & \text{if } y = 0. \end{cases}$$ ### Definition of Continuity: A function $$f(x, y)$$ is continuous at $$(0, 0)$$ if: $$\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0).$$ In this case, $$f(0, 0) = 0$$, so we need to check whether the limit of $$f(x, y)$$ as $$(x, y) \to (0, 0)$$ is 0. ### Step 1: Check the behavior along different paths. To investigate whether the limit exists, we'll check the limit along several paths as $$(x, y) \to (0, 0)$$. #### Path 1: $$y = 0$$ When $$y = 0$$, the function simplifies to: $$f(x, 0) = 0.$$ Thus, $$\lim_{x \to 0} f(x, 0) = 0.$$ #### Path 2: $$x = 0$$ When $$x = 0$$, the function simplifies to: $$f(0, y) = 0 \quad \text{for all } y \neq 0.$$ Thus, $$\lim_{y \to 0} f(0, y) = 0.$$ #### Path 3: $$y = x$$ When $$y = x$$, the function becomes: $$f(x, x) = \sin(x) \sin\left(\frac{1}{x}\right) \quad \text{for } x \neq 0.$$ Since $$\sin\left(\frac{1}{x}\right)$$ oscillates between -1 and 1 as $$x \to 0$$, the product $$\sin(x) \sin\left(\frac{1}{x}\right)$$ oscillates as well. Hence, the limit along this path does not exist. ### Step 2: Investigate general behavior as $$(x, y) \to (0, 0)$$ Consider the absolute value of the function for general $$(x, y) \neq (0, 0)$$: $$|f(x, y)| = |\sin(x) \sin\left(\frac{1}{y}\right)|.$$ Since $$|\sin(x)| \leq 1$$ and $$|\sin\left(\frac{1}{y}\right)| \leq 1$$, we have: $$|f(x, y)| \leq 1 \quad \text{for all } (x, y).$$ However, as we have seen from the path $$y = x$$, the function oscillates, and the limit does not settle to a single value as $$(x, y) \to (0, 0)$$. ### Step 3: Conclusion The function does not have a limit at $$(0, 0)$$ along all possible paths, particularly due to the oscillatory behavior along certain paths (e.g., $$y = x$$). Therefore, the function is **not continuous** at $$(0, 0)$$. Would you like further details or have any questions? Here are 5 questions related to this problem: 1. How would the behavior of the function change if we altered the definition for $$f(x, y)$$ when $$y = 0$$? 2. Can you provide another example of a function that is discontinuous at a specific point due to oscillation? 3. What does it mean for a function to be continuous on the entire domain $$\mathbb{R}^2$$? 4. How do the paths you choose affect the evaluation of limits in two-variable functions? 5. How would you apply the epsilon-delta definition of continuity to this problem? **Tip:** When evaluating limits for multivariable functions, always check behavior along multiple paths to ensure consistency of the limit.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Continuity in Two Variables
Limit along different paths
Formulas
f(x, y) = sin(x) * sin(1/y), if y ≠ 0
f(0, 0) = 0
Theorems
Definition of Continuity
Limit Evaluation by Pathwise Analysis
Suitable Grade Level
Undergraduate Mathematics
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