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Limit Of A Function From Definition

  1. Oct 20, 2009 #1
    I have to prove from the definiton of the limit of a function that the limit of f(x) as x tends to 2 equlas 3, given that;

    f(x)=(2x-1)

    I know that I have to find an epsilon such that |f(x)-l| [tex]\leq[/tex] [tex]\epsilon[/tex] and delta such that 0 [tex]\leq[/tex] |x-a| [tex]\leq[/tex] [tex]\delta[/tex]

    Nowing putting in the conditions for this f(x);

    |2x-4| [tex]\leq[/tex] [tex]\epsilon[/tex] and 0 [tex]\leq[/tex] |x-2| [tex]\leq[/tex] [tex]\delta[/tex]

    But I don't where to go from here. Any help would be great!
     
  2. jcsd
  3. Oct 20, 2009 #2

    arildno

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    Well, you have:

    |2x-4|=2|x-2|

    Can this be made arbitrarily small (i.e, below some fixed "epsilon"), as long as we pick a small enough "delta"?

    In particular, IF we want [itex]2|x-2|\leq\epsilon[/itex], what is the largest value we can pick for "delta" in order to ENSURE the validity of the above inequality?
     
  4. Oct 20, 2009 #3
    Often, we need to find [tex]\delta = \delta(\epsilon)[/tex] as a function of the [tex]\epsilon[/tex] we were given.
    That is, write [tex]|f(x)-l|[/tex] as a function [tex]|x-a|[/tex] and use your delta to show that it is less than epsilon
     
  5. Oct 20, 2009 #4
    So I know that |f(x)-l|=2|(x-2)|=2|x-a|

    Would it then be right to say that since

    |x-a|=|x-2| [tex]\leq[/tex] [tex]\delta[/tex]

    then |f(x)-l|=2|x-2| [tex]\leq[/tex] 2[tex]\delta[/tex]

    hence taking 2[tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex] would satisfy the problem??
     
  6. Oct 20, 2009 #5

    arildno

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    EXACTLY! :smile:

    (you meant, I think delta<=epsilon/2, I think..)
     
  7. Oct 20, 2009 #6
    Yeah that is what I meant! Woops. So is the following proof correct then;

    Definition of the Limit of a Function at a Point.
    Suppose that f(x) is defined on (a-R,a)U(a,a+R), for some R [tex]\succ[/tex] 0. Then f(x) tends to l as x tends to a if, given any [tex]\epsilon[/tex] in R[tex]^{+}[/tex] , there exists [tex]\delta[/tex] in R[tex]^{+}[/tex] such that,

    |f(x)-l| [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] |x-a| [tex]\prec[/tex] [tex]\delta[/tex]

    Now in this case f(x)=2x-1, l=3 and a=2

    Hence |2x-4| [tex]\prec[/tex] [tex]\epsilon[/tex]

    Now 2|x-2| [tex]\prec[/tex] [tex]\epsilon[/tex] = |x-a|[tex]\prec[/tex] [tex]\delta[/tex]

    Hence taking [tex]\delta[/tex] [tex]\leq[/tex] [tex]\frac{\epsilon}{2}[/tex]

    Then |f(x)-l| [tex]\prec[/tex] [tex]\epsilon[/tex] whenever 0 [tex]\prec[/tex] |x-a| [tex]\prec[/tex] [tex]\delta[/tex]

    As required
     
  8. Oct 20, 2009 #7

    arildno

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    Slightly scruffy-looking proof, but okay nonetheless! :smile:
     
  9. Oct 20, 2009 #8
    Thanks for the help!
     
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