print out - HW #12 : Due Thur, 12/4/08. Sec 5-1,p113: #10,...

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Unformatted text preview: 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 ( x ) is differentiable, but derivative is not continuous at x =0, ( ) ( ) 2 sin 1/ , , x x x f x x = = Pf : i) Let ( x ) be such a function, NTS: (x) is differentiable at x =0. Then, using definition of differentiability at a point: ( ) ( ) ( ) ( ) 2 1 1 sin ' 0 lim lim lim sin x x x f x f x x f x x x x-- -- = = = =- Hence, the limit exists and it is final f ( 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: '(x) is not continuous at x =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 cos(1/ x ) doesn't exist. Existence of limit of '( x )at x =0 is necessary continuity condition. #12 . Show that ( x ) is continuous but not differentiable at x =0. ( ) ( ) sin 1 / , , x x x f x x = = Pf : Let ( x ) be such a function. i) NTS: " >0 $ >0 : " x o D(), if | x-0|< f |( x )-(0)|< ( ) ( ) 1 1 1 sin sin sin f x f x x x x x x x ---- =- = = = , since |sin(1/ x )| 1 " x 0 Simply, choose = . Hence, ( x ) is continuous at x =0. ii) NTS: ( x ) fails the definition of differentiability at x =0 ( ) ( ) ( ) 1 sin 1 ' lim lim limsin x x x f x f x x f x x x x- -- = = =- doesn't have a limit at...
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print out - HW #12 : Due Thur, 12/4/08. Sec 5-1,p113: #10,...

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