31. (a)Note that 2x #sin }1x}#x, for all x,so limx→01xsin }1x}250 by the Sandwich Theorem.Therefore,fis continuous at x50.(b)}f(01hh)2f(0)}55sin }1h}(c)The limit does not exist because sin }1h}oscillatesbetween 21 and 1 an infinite number of timesarbitrarily close toh 50 (that is, for hin any openinterval containing 0).(d)No, because the limit in part (c) does not exist.(e)}g(01hh)2g(0)}55hsin }1h}As noted in part (a), the limit of this as xapproacheszero is 0, so g9(0)50.■Section 3.3Rules for Differentiation(pp. 112–121)Quick Review 3.31.(x222)(x2111)5x2x211x2?122x2122 ?15x1x222x21222.1}x21x1}2215}x21x1}5}xx2}1}1x}5x1x213.3x22}2x}1}x52}53x222x2115x224.}3x4222xx2314}5}32xx42}2}22xx32}1}24x2}5}32}x22x12x225.(x2112)(x2211)5x21x221x21?112x2212 ?15x231x2112x22126.
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