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Converging sequence problem

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Given a sequence [itex](x_n), x_n > 0[/itex] for every [itex]n\in\mathbb{N}[/itex] and [itex]\lim\limits_{n\to\infty} x_n = L > 0[/itex], show that [itex]\ln x_n\to \ln L[/itex] when [itex]n\to\infty[/itex].

    2. Relevant equations


    3. The attempt at a solution
    As logarithm function is an elementary function, meaning it is continous in its domain [itex]D:= (0,\infty)[/itex] then we have that:
    [itex]\forall\varepsilon >0, \exists\delta >0:\forall x\in D\left ( 0<|x-L|<\delta\Rightarrow |\ln x - \ln L|<\varepsilon\right )[/itex]
    Provided that [itex]x_n \to L[/itex], then there exists [itex]N\in\mathbb{N}[/itex] such that:
    [itex]n\geq N\Rightarrow |x_n-L|<\delta[/itex], from which it follows that:
    [itex](n\geq N\Rightarrow |\ln x_n -\ln L|<\varepsilon) \Leftrightarrow \lim\limits_{n\to\infty} \ln x_n =\ln L[/itex]

    I googled this problem, but I couldn't find an epsilon-delta argument, so I gave it a go. Is this convincing?
     
    Last edited: Nov 3, 2015
  2. jcsd
  3. Nov 3, 2015 #2

    Ray Vickson

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    Are you allowed to use the fact (as you have done) that ##\ln x## is continuous for ##x > 0##, or are you essentially required to prove that first?
     
  4. Nov 3, 2015 #3
    Define [itex]f:\mathbb{R}\to\mathbb{R}[/itex] as [itex]f(x) = a^x, a >0[/itex]. We claim that [itex]f(x)[/itex] is continous in [itex]\mathbb{R}[/itex].
    Fix [itex]z\in\mathbb{R}[/itex]. If [itex]x\to z[/itex], then [itex]x-z\to 0[/itex] and [itex]\lim\limits_{x\to z} a^{x-z} = 1[/itex] from which:
    [itex]\lim\limits_{x\to z}a^x = \lim\limits_{x\to z} a^za^{x-z} = a^z\lim\limits_{x\to z}a^{x-z} = a^z[/itex].
    We have established that [itex]f[/itex] is continous in [itex](-\infty ,\infty)[/itex]

    Bolzano-Cauchy theorem: Let [itex]f[/itex] be continous in [itex]D\subseteq\mathbb{R}[/itex]. If [itex]y_1, y_2[/itex] are two different values of the function, then every [itex]y[/itex] between [itex]y_1[/itex] and [itex]y_2[/itex] is also a value of the function.
    Therefore the set of values of the function [itex]f[/itex] that is continous in some interval, is also an interval.

    We have a function [itex]f: (-\infty, \infty)\to (0,\infty )[/itex], which is continous. Its inverse (implies it is invertible - bijection, which it is if [itex]a>1[/itex] or [itex]a<1[/itex]) is defined to be the logarithm function: [itex]f^{-1}: (0, \infty)\to (-\infty,\infty)[/itex].
    [itex]f^{-1}: = \log _a (x)[/itex] is also continous ([itex]f[/itex] is strictly monotone for [itex]a\neq 1[/itex], its inverse is also strictly monotone.)
     
    Last edited: Nov 3, 2015
  5. Nov 3, 2015 #4
    Google 'sequential definition of continuity'.
     
  6. Nov 3, 2015 #5
    English -.- We call the same thing as "Heine's criterion for continuity"
     
  7. Nov 3, 2015 #6
    And yes your proof looks OK to me except that you must mention that ##L>0## and also here

    [itex]\forall\varepsilon >0, \exists\delta >0:\forall x\in D ... [/itex]
     
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