Math Problem Statement
Show that the iterated limits lim (x->0) (lim (y->0) f(x,y)) and lim (y->0) (lim (x->0) f(x,y)) do not exist, but the two-variable limit lim (x,y)->(0,0) f(x,y) = 0, where f(x, y) is defined as: f(x, y) = (x + y)^2 cos(1/x) cos(1/y) if xy ≠ 0 and f(x, y) = 0 if xy = 0.
Solution
Let's analyze the given mathematical problem:
Problem Summary:
We are given a function defined as:
\begin{cases} (x + y)^2 \cos\left(\frac{1}{x}\right) \cos\left(\frac{1}{y}\right), & \text{if } xy \neq 0, \\ 0, & \text{if } xy = 0. \end{cases}$$ The goal is to: 1. Show that the iterated limits: $$\lim_{x \to 0} \left( \lim_{y \to 0} f(x, y) \right) \quad \text{and} \quad \lim_{y \to 0} \left( \lim_{x \to 0} f(x, y) \right)$$ do **not exist**. 2. Prove that the two-variable limit: $$\lim_{(x, y) \to (0, 0)} f(x, y) = 0.$$ --- ### Step 1: Iterated Limits Let’s examine $$\lim_{y \to 0} f(x, y)$$ first, keeping $$x$$ fixed. #### Case 1: $$y \to 0$$ while $$x \neq 0$$: $$f(x, y) = (x + y)^2 \cos\left(\frac{1}{x}\right) \cos\left(\frac{1}{y}\right).$$ Here: - $$\cos\left(\frac{1}{y}\right)$$ oscillates between $$-1$$ and $$1$$ as $$y \to 0$$. - Thus, the product $$f(x, y)$$ does not settle to a single value because of the oscillatory nature of $$\cos\left(\frac{1}{y}\right)$$. Therefore: $$\lim_{y \to 0} f(x, y) \quad \text{does not exist}.$$ #### Case 2: $$x \to 0$$ after $$y \to 0$$: Since the inner limit $$\lim_{y \to 0} f(x, y)$$ does not exist, the iterated limit $$\lim_{x \to 0} \left( \lim_{y \to 0} f(x, y) \right)$$ does not exist either. A similar argument holds for the reverse order of the limits: $$\lim_{y \to 0} \left( \lim_{x \to 0} f(x, y) \right).$$ Thus, neither of the iterated limits exists. --- ### Step 2: Two-Variable Limit as $$(x, y) \to (0, 0)$$ Now, we analyze: $$f(x, y) = (x + y)^2 \cos\left(\frac{1}{x}\right) \cos\left(\frac{1}{y}\right).$$ #### Case 1: $$xy = 0$$: From the definition, $$f(x, y) = 0$$. #### Case 2: $$xy \neq 0$$: - The term $$(x + y)^2$$ approaches $$0^2 = 0$$ as $$(x, y) \to (0, 0)$$. - The oscillatory terms $$\cos\left(\frac{1}{x}\right)$$ and $$\cos\left(\frac{1}{y}\right)$$ remain bounded between $$-1$$ and $$1$$. Thus, the product: $$f(x, y) = (x + y)^2 \cos\left(\frac{1}{x}\right) \cos\left(\frac{1}{y}\right)$$ approaches $$0 \cdot (\text{bounded terms}) = 0$$ as $$(x, y) \to (0, 0)$$. Therefore: $$\lim_{(x, y) \to (0, 0)} f(x, y) = 0.$$ --- ### Conclusion: 1. The iterated limits do **not exist** due to the oscillatory behavior of $$\cos\left(\frac{1}{x}\right)$$ and $$\cos\left(\frac{1}{y}\right)$$. 2. The two-variable limit $$\lim_{(x, y) \to (0, 0)} f(x, y) = 0$$. Would you like a detailed explanation of any step or further examples? --- ### Related Questions: 1. What are iterated limits, and how do they differ from two-variable limits? 2. How can oscillatory behavior affect the existence of limits? 3. Can you provide a different example where iterated limits fail to exist, but the two-variable limit does? 4. What happens if we approach $$(0, 0)$$ along specific paths like $$y = mx$$ in this function? 5. How would the result change if $$\cos\left(\frac{1}{x}\right)$$ was replaced with a bounded non-oscillatory function? --- ### Tip: When dealing with limits of functions of two variables, always check for path-dependence and oscillatory terms to understand the behavior around the point of interest.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits of functions of two variables
Iterated limits
Oscillatory behavior in limits
Formulas
f(x, y) = (x + y)^2 cos(1/x) cos(1/y) if xy ≠ 0; f(x, y) = 0 if xy = 0
Theorems
Properties of limits in multivariable calculus
Suitable Grade Level
Undergraduate Mathematics
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