PracticeExam1-soln

# PracticeExam1-soln - Solutions to practice final1 1(a...

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Unformatted text preview: Solutions to practice final1 1(a). Answer : Apply two path test. Let y = mx 2 , m ̸ = 1 and m ≥ 0. Then lim ( x, y ) → (0 , 0) xy 1 / 2 x 2 − y = lim ( x, y ) → (0 , 0) x · √ mx x 2 − mx 2 = √ m 1 − m . If m = 4, Limit = − 2 3 . If m = 9, Limit = − 3 8 . So by two path test, the limit does not exist . 1(b). Answer : F ′ ( x ) = f ( e x 2 )( e x 2 ) ′ = f ( e x 2 ) · e x 2 · 2 x. Then F ′ (1) = f ( e 1 ) · e · 2 = 4 · e · 2 = 8 e . 1(c). Answer : (A) describes an elliptic paraboloid. For 0, the equation of level curve is x 2 4 + y 2 9 = 0, it is just the point (0 , 0). For 1, the equation of level curve is x 2 4 + y 2 9 = 1, it is a ellipse. There is no level curve corresponding − 1. 1(d). Answer : Let u = x 2 , then du = 2 x dx . Then ∫ xe − x 2 dx = ∫ 1 2 e − u du = − 1 2 e − u + C = − 1 2 e − x 2 + C . Thus ∫ ∞ xe − x 2 dx = lim b →∞ ∫ b xe − x 2 dx = lim b →∞ − 1 2 e − x 2 | b = lim b →∞ − 1 2 [ e − b 2 − e ] = 1 2 ....
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PracticeExam1-soln - Solutions to practice final1 1(a...

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