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# s5 - Solution to Set 5 1 We proceed by showing that all...

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Solution to Set 5 1. We proceed by showing that all points not in L are exterior points of L , so that L contains all the boundary points and is a close set. Let ( x, y ) be a point not in L , then by the definition of L , x negationslash = 0. Pick an epsilon1 which has the value | x | , then all the points inside the open disk D epsilon1 ( x, y ) are not in L (( a, b ) D epsilon1 ( x, y ) | x - a | < epsilon1 = | x | a negationslash = 0). Thus ( x, y ) is an exterior point of L . 2. First, it is easy to show that f x (0 , 0) = f y (0 , 0) = f (0 , 0) = 0. Then define epsilon1 1 ( x, y ) = y sin( x/y ) and epsilon1 2 ( x, y ) = 0. In a small disk around (0 , 0), Δ x = x , Δ y = y , and Δ f = f ( x, y ) - f (0 , 0) can be written in the form Δ f = f x (0 , 0)Δ x + f y (0 , 0)Δ y + epsilon1 1 Δ x + epsilon1 2 Δ y The limit lim ( x,y ) (0 , 0) epsilon1 1 ( x, y ) is 0 as sin( x/y ) is bounded by ± 1 and lim ( x,y ) (0 , 0) y = 0. Therefore the definition of differentiability is satisfied at (0 , 0). 3. (a) ∂u w = - 1 v sin u v ∂v w = u v 2 sin u v 2 ∂u 2 w = - 1 v 2 cos u v 2 ∂v 2 w = - u v 3 parenleftBig u v cos u v + 2 sin

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