This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 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...
View Full Document
This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.
- Fall '09
- Multivariable Calculus