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Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine a along the curve y = x². When x is positive, we have xy lim (x, y) → (0,0) √ x² + y2 When x is negative, we have xy lim (x, y) (0, 0) √ x² + y² = lim X→0 = lim X-0 0 x2 +x4 = lim X-0 lim X-0-1 x²

Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine a along the curve y = x². When x is positive, we have xy lim (x, y) → (0,0) √ x² + y2 When x is negative, we have xy lim (x, y) (0, 0) √ x² + y² = lim X→0 = lim X-0 0 x2 +x4 = lim X-0 lim X-0-1 x²

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Title: Exploring Limits of Multivariable Functions** **Find the limit, if it exists:** \[ \lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} \] **Step 1:** To examine \(\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}\), first approach \((0, 0)\) along the x-axis. On this path, all points have \(y = 0\). **Step 2:** Since \(y = 0\) for all points on this path, then we have: \[ \lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{x \to 0} \frac{x \cdot 0}{\sqrt{x^2 + 0}} = 0 \] **Step 3:** Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach along the curve \(y = x^2\).
Transcribed Image Text:**Title: Exploring Limits of Multivariable Functions** **Find the limit, if it exists:** \[ \lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} \] **Step 1:** To examine \(\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}\), first approach \((0, 0)\) along the x-axis. On this path, all points have \(y = 0\). **Step 2:** Since \(y = 0\) for all points on this path, then we have: \[ \lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{x \to 0} \frac{x \cdot 0}{\sqrt{x^2 + 0}} = 0 \] **Step 3:** Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine another approach along the curve \(y = x^2\).
**Step 2** Since \( y = 0 \) for all points on this path, then we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot 0}{\sqrt{x^2 + 0}} \] which simplifies to \[ = 0 \] --- **Step 3** Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine approaching along the curve \( y = x^2 \). When \( x \) is positive, we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot x^2}{\sqrt{x^2 + x^4}} = \lim_{{x \to 0}} \frac{x^3}{\sqrt{x^2 + x^4}} \] --- When \( x \) is negative, we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot x^2}{\sqrt{x^2 + x^4}} = \lim_{{x \to 0}} \frac{x^3}{\sqrt{x^2 + x^4}} \] --- The steps describe finding limits along different paths, demonstrating the dependence of limits on the chosen path.
Transcribed Image Text:**Step 2** Since \( y = 0 \) for all points on this path, then we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot 0}{\sqrt{x^2 + 0}} \] which simplifies to \[ = 0 \] --- **Step 3** Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine approaching along the curve \( y = x^2 \). When \( x \) is positive, we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot x^2}{\sqrt{x^2 + x^4}} = \lim_{{x \to 0}} \frac{x^3}{\sqrt{x^2 + x^4}} \] --- When \( x \) is negative, we have \[ \lim_{{(x, y) \to (0, 0)}} \frac{xy}{\sqrt{x^2 + y^2}} = \lim_{{x \to 0}} \frac{x \cdot x^2}{\sqrt{x^2 + x^4}} = \lim_{{x \to 0}} \frac{x^3}{\sqrt{x^2 + x^4}} \] --- The steps describe finding limits along different paths, demonstrating the dependence of limits on the chosen path.
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