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Epsilon delta application

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Let [tex]f: \Re \rightarrow \Re [/tex] and [tex]g: \Re \rightarrow \Re [/tex] be functions such that
    [tex] lim_{x \rightarrow 1} f(x)=\alpha [/tex]
    and
    [tex] lim_{x \rightarrow 1} g(x)=\beta [/tex]
    for some [tex] \alpha, \beta \in \Re [/tex] with [tex] \alpha < \beta [/tex]. Use the [tex] \epsilon-\delta [/tex] definition of a limit to prove there exists a number [tex]\delta >0[/tex]
    such that [tex] f(x)<g(x) [/tex] for all [tex] x [/tex] satisfying [tex]1- \delta < x< 1+ \delta [/tex].


    2. Relevant equations


    3. The attempt at a solution

    By definition I know that there exists a
    [tex] \delta_1 >0 [/tex] s.t. [tex] \left|f(x)-\alpha \right|< \epsilon_1[/tex] [tex]\forall \left| x-\alpha \right| < \delta_1 [/tex]
    and
    [tex] \delta_2 >0 [/tex] s.t. [tex] \left|g(x)-\beta \right|< \epsilon_2[/tex] [tex]\forall \left| x-\beta \right| < \delta_2 [/tex].

    Then I know that the [tex]1- \delta < x< 1+ \delta [/tex] can simplify to [tex] \left| x-1 \right| < \delta [/tex].

    Perhaps I should set [tex] \delta = min( \delta_1, \delta_2) [/tex]? It seems that I need to show that if I can get [tex] x [/tex] close enough to 1 ([tex] \left| x-1 \right| < \delta [/tex]) then [tex] f(x) [/tex] can get close enough to [tex] \alpha [/tex] and [tex] g(x) [/tex] can get close enough to [tex]\beta[/tex], thus because [tex]\alpha < \beta[/tex] we have [tex]f(x) < g(x)[/tex]. Am I on the right track? If so how would I start to prove this?


    Thanks!
     
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2
    You will need to find [itex]\delta[/itex], but remember that [itex]\delta[/itex] depends on [itex]\epsilon[/itex]. You're on the right track; what do you think you need to pick for epsilon so that f(x) is "close enough" to [itex]\alpha[/itex] and g(x) is "close enough" to [itex]\beta[/itex]?
     
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