math237_a2

math237_a2 - SOLUTIONS FOR ASSIGNMENT 2 A3 Consider f R 2...

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Unformatted text preview: SOLUTIONS FOR ASSIGNMENT 2 A3. Consider f : R 2 → R defined by f ( x,y ) = sin( xy ) ln( x 2 + y 2 + 1) , ( x,y ) negationslash = (0 , 0) f (0 , 0) = . Prove that f x (0 , 0) and f y (0 , 0) exist, but that f is not continuous at (0 , 0) . From the definition of first order partial derivatives we have: f x (0 , 0) = lim h → f (0 + h, 0) − f (0 , 0) h = lim h → sin 0 ln( h 2 +1) − h = 0 , and, similarly, f y (0 , 0) = lim h → f (0 , 0 + h ) − f (0 , 0) h = lim h → sin 0 ln( h 2 +1) h = 0 . Since the above limits exist we conclude that f x (0 , 0) and f y (0 , 0) exist. In order to show that f is not continuous at (0 , 0) we need to show that lim ( x,y ) → (0 , 0) sin( xy ) ln( x 2 + y 2 + 1) negationslash = 0 . We evaluate this limit along the straight line paths y = mx, ∀ m ∈ R . We have: lim x → sin( mx 2 ) ln((1 + m 2 ) x 2 + 1) = m 1 + m 2 lim x → ((1 + m 2 ) x 2 + 1) cos ( mx 2 ) (by l ′ Hospital ′ s rule) = m 1 + m 2 , ∀ m ∈ R . Since the above limit depends on the slope m of the straight line paths, it follows from the two paths’ rule that the limit lim ( x,y ) → (0 , 0) sin( xy ) ln( x 2 + y 2 + 1) does not exist and hence it cannot equal 0 = f (0 , 0) . Thus f is not continuous at (0 , 0) and this 1 ends the proof. B1 (v). For the function f ( x,y ) = x 2 − 6 y 2 | x | + 3 | y | determine whether or not lim ( x,y ) → (0 , 0) f ( x,y ) exists and, if possible, define f (0 , 0) so as to make the...
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This note was uploaded on 07/28/2009 for the course MATH MATH137 taught by Professor Oancea during the Spring '08 term at Waterloo.

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math237_a2 - SOLUTIONS FOR ASSIGNMENT 2 A3 Consider f R 2...

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