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# print out - HW#12 Due Thur Sec 5-1,p113#10 12 Sec...

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HW #12 : Due Thur, 12/4/08. Sec 5-1,p113: #10, 12. Sec 5-2,p.129:#2, 5, 8, 12 #10. Show that ƒ( x ) is differentiable, but derivative is not continuous at x =0, ( ) ( ) 2 sin 1/ , 0 0 , 0 x x x f x x = = Pf : i) Let ƒ( x ) be such a function, NTS: ƒ(x) is differentiable at x =0. Then, using definition of differentiability at a point: ( ) ( ) ( ) ( ) 2 1 1 0 0 0 0 sin 0 ' 0 lim lim lim sin 0 0 x x x f x f x x f x x x x - - - - = = = = - Hence, the limit exists and it is final fl ƒ( x ) is differentiable at x =0, by definition. ii) From elementary calculus: ( ) 2 2 2 1 1 1 1 1 1 ' sin 2 sin cos 2 sin cos d f x x x x x dx x x x x x x = = + - = - NTS: ƒ'(x) is not continuous at x =0 ( ) 0 0 0 0 1 1 1 1 lim ' lim 2 sin cos lim 2 sin lim cos x x x x f x x x x x x x = - = - doesn't exist because lim x 0 cos(1/ x ) doesn't exist. Existence of limit of ƒ'( x )at x =0 is necessary continuity condition. É #12 . Show that ƒ( x ) is continuous but not differentiable at x =0. ( ) ( ) sin 1/ , 0 0 , 0 x x x f x x = = Pf : Let ƒ( x ) be such a function. i) NTS: " ε >0 \$ δ >0 : " x oe D(ƒ), if | x -0|< δ fl |ƒ( x )-ƒ(0)|< ε ( ) ( ) 1 1 1 0 sin 0 sin sin f x f x x x x x x x δ - - - - = - = = = , since |sin(1/ x )| 1 " x 0 Simply, choose δ = ε . Hence, ƒ( x ) is continuous at x =0.

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