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An^bn as an->a and bn->b

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    an -> a
    bn -> b
    prove that anbn = ab
    Its all Sequence of course
  2. jcsd
  3. Nov 11, 2009 #2
    Hint: [tex] (a_n)^{b_n} = e^{b_n log(a_n)} [/tex]

    Now use continuity of the exponential and logarithm function to take the limits "inside the function".
  4. Nov 11, 2009 #3
    By the way, we don't want to prove that [tex] (a_n)^{b_n} = a^b [/tex]. We want to prove that [tex] (a_n)^{b_n} \rightarrow a^b [/tex] as [tex] n \rightarrow \infty [/tex]
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