21-356 Lecture 3 - and assume that x0 is an accumulation...

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Example 5 Let f ( x, y ) := ! 1 if y = x 2 , x % = 0 , 0 otherwise. Given a direction v = ( v 1 , v 2 ) , the line L through 0 in the direction v intersects the parabola y = x 2 only in 0 and in at most one point. Hence, if t is very small, f (0 + tv 1 , 0 + tv 2 ) = 0 . It follows that ! f ! v (0 , 0) = lim t ! 0 f (0 + tv 1 , 0 + tv 2 ) " f (0 , 0) t = lim t ! 0 0 " 0 t = 0 . However, f is not continuous in 0 , since f " x, x 2 # = 1 & 1 as x & 0 , while f ( x, 0) = 0 & 0 as x & 0 . Exercise 6 Let f ( x, y ) := $ x 2 y x 4 + y 2 if ( x, y ) % = (0 , 0) , 0 if ( x, y ) = (0 , 0) . Find all directional derivatives of f at 0 and prove that f is not continuous at 0 . 5
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The previous examples show that in dimension N ! 2 partial derivatives do not give the same kind of results as in the case N = 1 . To solve this problem, we introduce a stronger notion of derivative, namely, the notion of di ! erentiability. We recall that a function T : R N " R is linear if T ( x + y ) = T ( x ) + T ( y ) for all x , y # R N and T ( s x ) = sT ( x ) for all s # R and x # R N . Write x = ! N i =1 x i e i . Then by the linearity of T , T ( x ) = T " N # i =1 x i e i $ = N # i =1 x i T ( e i ) .
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