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Math Help - real sequences analysis help

  1. #1
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    real sequences analysis help

    1. assume that an->2 and bn-->3 evaluate lim x->inf (2an -3bn)
    a) -5 (b)0 (c) the limit doesnot exist (d) the limit cannot be determined

    2. Assume that an-->2 and bn-->3 Evaluate lim x->inf (an-2)/bn
    (a)0 (b) the limit doesnot exist (c) the limit canot be determined

    3.We know from trignometrythat |sin(n)<=1 for all n.This shows that 0<= |sin(n)/n|<= 1/n and hence that -1/n <=|sin(n)/n <= 1/n. We can also evaluate the lim x->inf (sin(n)/n by using
    (a) product rule for limit of sequences.
    (b) quotient rule for limits of sequences.
    (c) the squeeze theorem for sequence

    (4) find lim x-->inf sin(n)/n
    (a)1 (b) 0 (c) limit doesnot exist
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  2. #2
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    Quote Originally Posted by bobby77 View Post
    1. assume that an->2 and bn-->3 evaluate lim x->inf (2an -3bn)
    Use the theorem about sequences, i.e. direct substitution.
    2(2)-3(3)=4-9=-5

    2. Assume that an-->2 and bn-->3 Evaluate lim x->inf (an-2)/bn
    Since lim bn does not equal to zero. We can use division rule.
    (2-2)/3=0

    3.We know from trignometrythat |sin(n)|<=1 for all n.This shows that 0<= |sin(n)/n|<= 1/n and hence that -1/n <=|sin(n)/n| <= 1/n. We can also evaluate the lim x->inf (sin(n)/n by using
    My favorite, the "squeeze theorem".
    (Other favorite is composite function rule).
    (4) find lim x-->inf sin(n)/n
    We have that,
    |sin(n)/n|<1/n
    Thus,
    -1/n<sin(n)/n<1/n
    Squeeze theorem.
    Both -1/n and 1/n ---> 0
    Thus, limit is zero.
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