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Unformatted text preview: and denominator: lim x e x-e-x sin 5 x , lim x sin 4 x 1-cos( x 2 ) . To hand in: (1) Problem 66 on page 63 of the text. Note: There is a typo: s q = q k =0 1 k ! (2) (a) Find the 4th order Taylor polynomial (about a = 0 of cos(sin x ). (b) Use the result of (a) (not lHospitals rule) to compute lim x cos(sin x )-cos x x 2 sin( x 2 ) . (3) Find all values of x for which the followins series (a) converges absolutely, (b) condition-ally f ( x ) = X n =0 (-1) n ( x-3) n n + 2 . 1...
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- Fall '08