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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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Find the limit, if it exists. Step 1 xy √x² + y² lim (x, y) → (0, 0) V To examine xy lim (x, y) (0, 0) √x² + y²' On this path, all points have y = 0 first approach (0, 0) along the x-axis. xy lim (x,y) → (0, 0) Vx² + y2 Step 2 Since y = 0 for all points on this path, then we have = lim X 0 Y = 0 X. 0 √x² +0 0 0 45 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 appr along the curve y = x²,
Transcribed Image Text:Find the limit, if it exists. Step 1 xy √x² + y² lim (x, y) → (0, 0) V To examine xy lim (x, y) (0, 0) √x² + y²' On this path, all points have y = 0 first approach (0, 0) along the x-axis. xy lim (x,y) → (0, 0) Vx² + y2 Step 2 Since y = 0 for all points on this path, then we have = lim X 0 Y = 0 X. 0 √x² +0 0 0 45 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 appr along the curve y = x²,
Step 2 Since y = 0 for all points on this path, then we have xy lim = lim (x, y) (0, 0) √x² + y² X 0 lim (x,y) → (0,0) √x² + y2 When x is negative, we have Submit xy lim (x, y) (0, 0) V x² + y² Step 3 Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine ar along the curve y = x². When x is positive, we have = lim = X→0 lim X→ 0 XO Skip (you cannot come back) √x² + x² + 0 x² + x4 = lim X→ 0 = lim x → 0 0
Transcribed Image Text:Step 2 Since y = 0 for all points on this path, then we have xy lim = lim (x, y) (0, 0) √x² + y² X 0 lim (x,y) → (0,0) √x² + y2 When x is negative, we have Submit xy lim (x, y) (0, 0) V x² + y² Step 3 Similarly, approaching along the y-axis yields a limit equal to 0. Since these two limits are the same, we will examine ar along the curve y = x². When x is positive, we have = lim = X→0 lim X→ 0 XO Skip (you cannot come back) √x² + x² + 0 x² + x4 = lim X→ 0 = lim x → 0 0
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