Calc 3 Exam 2 Sin - CALCULUS AND ANALYTIC GEOMETRY III...

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CALCULUS AND ANALYTIC GEOMETRY III SOLUTIONS TO SAMPLE TEST 2 Please let me know if you find errors or typos. 1. The limit does not exist, for if it did, then we would obtain the same limit when approaching (0 , 0) along any curve. However, this is not so, since if we approach along the line y = x 3 , we have lim ( x,x 3 ) (0 , 0) x 3 .x 3 ( x 3 ) 2 + x 6 = lim x 0 x 6 2 x 6 = 1 2 , while for the line x = 0, we have lim (0 ,y ) (0 , 0) 0 3 .y 0 6 + y 2 = lim y 0 0 = 0 . Since the limit of f ( x, y ) does not exist at (0 , 0), no choice of the value of f (0 , 0) will result in a function which is continuous at (0 , 0), since for continuity, the limit must exist in the first place. 2. f x ( x, y ) = 4 x 3 - 4 y , f y ( x, y ) = 4 y 3 - 4 x . These partial derivatives exist every- where, so the only critical points are the points ( x, y ) where both partials are zero. Thus we must solve the simultaneous equations: (1) ± x 3 - y = 0 y 3 - x = 0 . Substituting, we get
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Calc 3 Exam 2 Sin - CALCULUS AND ANALYTIC GEOMETRY III...

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