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Homework Help: Sequences of positive numbers and limits

  1. Jul 31, 2005 #1
    Let (x(n)) and (y(n)) be sequences of positive numbers such that lim(x(n)/y(n)) = 0.
    If lim(x(n)) = +∞, then lim(y(n)) = +∞
    If (y(n)) is bounded, then lim(x(n)) = 0

    To me this is self-evident. But HOW can it be proved?
     
  2. jcsd
  3. Jul 31, 2005 #2
    applications of definitions:

    for any [tex]\epsilon > 0[/tex] there's an [tex]n_\epsilon : [/tex] if [tex]n > n_\epsilon[/tex] then [tex]|\frac {x_n} {y_n}| < \epsilon [/tex] and for any [tex]M > 0[/tex] there's an [tex]n_M : [/tex] if [tex]n > n_M[/tex] then [tex]|x_n| > M [/tex]

    then for any [tex]M > 0[/tex] there's a k := max {[tex]n_M; n_\epsilon[/tex]} : if [tex]n > k[/tex] then [tex]|\frac {x_n} {y_n}| < \epsilon[/tex] and [tex]|x_n|> M[/tex] then [tex] \epsilon|y_n|>|x_n|> M[/tex] then [tex]|y_n|> \frac {M} {\epsilon} > M[/tex]

    CVD
     
    Last edited: Jul 31, 2005
  4. Jul 31, 2005 #3
    The other is possibly simpler:

    [tex]\exists M > 0 : \forall n \in \mathbb{N}, |y_n| < M[/tex]
    [tex]
    \forall \epsilon > 0, \exists n_{\epsilon} : n > n_{\epsilon} \Rightarrow |\frac {x_n} {y_n}| < \epsilon [/tex]
    [tex]\Rightarrow \forall \delta > 0, \exists n_{\delta} : n > n_{\delta} \Rightarrow \frac {|x_n|} {\delta} < |y_n| < M [/tex]
    [tex]\Rightarrow |x_n| < \delta M, \delta := \frac {\epsilon} {M} [/tex]
    [tex]\Rightarrow \exists k := max(n_\epsilon; n_\delta) : n > k \Rightarrow |x_n| < \epsilon[/tex]

    CVD
     
    Last edited: Jul 31, 2005
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