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Unformatted text preview: ± ± ( f ( x + h )f ( x )) f ( x ) h f ( x ) 2 f ( x + h ) ± ± ± ± ≤ o (  h  )  h  1  f ( x ) f ( x + h )  +  ( f ( x + h )f ( x )   f ( x )   f ( x ) 2 f ( x + h )  Since f is continuous, we have f ( x + h )f ( x ) → 0, and by deﬁnition o (  h  ) /  h  → 0, as  h  → 0, so we have 1  h  ± ± ± ± 1 f ( x + h )1 f ( x ) + f ( x ) h f ( x ) 2 ± ± ± ± → 0 as  h  → , which means 1 /f is diﬀerentiable at x with derivative ² 1 f ³ ( x ) =f ( x ) f ( x ) 2 . (b) Let T : R n → R m be the 0 mapping. Then since g (0) must be 0, for h ∈ U we have 1  h   g (0 + h )g (0)T h  =  g ( h )   h  ≤  h  α1 → 0 as  h  → . This shows that g (0) = T = 0. 2...
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 Fall '07
 Alexander
 Math, ax, dx dy dz

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