yZqAKX-03

yZqAKX-03 - ASSIGNMENT 3 SOLUTIONS MAT 473 SPRING 2012...

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ASSIGNMENT 3 · SOLUTIONS MAT 473 · SPRING 2012 Exercise 4.2. Suppose f : R 2 R is defned by f ( x, y )= xy 2 x 2 + y 2 iF ( x, y ) ° =(0 , 0) 0i F ( x, y )=(0 , 0) . Show that all partial derivatives oF f exist everywhere, and f is continuous everywhere, but f ° (0 , 0) still does not exist. Proof. Continuity at ( x, y ) ° =(0 , 0) is clear, since f is defned as a rational Function at such points. ±or continuity at (0 , 0), note that For ( x, y ) ° =(0 , 0), | f ( x, y ) | = | x | y 2 x 2 + y 2 ≤| x | , so f ( x, y ) 0as( x, y ) 0. ±or the partial derivatives, again there is no issue at ( x, y ) ° =( 0 , 0). ±or the partial derivatives at (0 , 0), compute directly: D 1 f (0 , 0) = lim t 0 f (0 + t, 0) f (0 , 0) t =l im t 0 0 t =0 and D 2 f (0 , 0) = lim t 0 f (0 ,t ) f (0 , 0) t =l im t 0 0 t =0 . Thus, iF f were di²erentiable at (0 , 0), we’d have [ f ° (0 , 0)] = ° 00 ± , which would imply that f ° (0 , 0) = 0, which in turn would imply that lim ( h,k ) (0 , 0) f (0 + h, 0+ k ) f (0 , 0) 0( h, k ) ± ( h, k ) ± =0 . However, this is not the case, since For example lim ( h,h ) (0 , 0) f (0 + h, 0+ h ) f (0 , 0) 0( h, h ) ± ( h, h ) ± =l im h
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yZqAKX-03 - ASSIGNMENT 3 SOLUTIONS MAT 473 SPRING 2012...

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