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Delta epsilon limits

  1. Mar 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Given that lim f(x)=L as x approaches a , prove that lim (x+f(x))=a+L as x approaches a View attachment 9630 . Your proof cannot assume that the limit of a sum of two functions is the sum of their individual limits. You must use the delta-epsilon definition of limit in your proof.
    and
    Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a

    2. Relevant equations



    3. The attempt at a solution
    attempt is in the attatchment
     

    Attached Files:

  2. jcsd
  3. Mar 29, 2007 #2

    VietDao29

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    Homework Helper

    That part is wrong, you cannot sub L for f(x) like that, it's not correct.

    Now, back at the start, the problem gives that:
    [tex]\lim_{x \rightarrow a} f(x) = L[/tex], and you must use this to prove: [tex]\lim_{x \rightarrow a} [x + f(x) ] = a + L[/tex]
    [tex]\lim_{x \rightarrow a} f(x) = L[/tex] means that, for any arbitrary small [tex]\epsilon[/tex], there exists a [tex]\delta[/tex], such that:
    [tex]0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon[/tex]


    What we should prove is:
    For any arbitrary small [tex]\epsilon_1[/tex], there exists a [tex]\delta_1[/tex], such that:
    [tex]0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < \epsilon_1[/tex]

    We will deal with this part:
    [tex]|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L|[/tex]


    Now we choose, [tex]\epsilon = \frac{\epsilon_1}{2}[/tex].

    Now, there should exists a [tex]\delta[/tex], such that: [tex]0 < |x - a| < \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}[/tex]
    Choose [tex]\delta_1[/tex], so that: [tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right)[/tex]

    [tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \delta[/tex], so we have:
    [tex]0 < |x - a| < \delta_1 \leq \delta \Rightarrow |f(x) - L| < \epsilon = \frac{\epsilon_1}{2}[/tex]

    [tex]\delta_1 = \mbox{min} \left( \delta, \frac{\epsilon_1}{2} \right) \Rightarrow \delta_1 \leq \frac{\epsilon}{2}[/tex]


    Now, if we have:
    [tex]0 < |x - a| < \delta_1 \Rightarrow |[x + f(x)] - [a + L]| < |x - a| + |f(x) - L| < \delta_1 + \epsilon < \frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1[/tex], so:
    [tex]\lim_{x \rightarrow a} [x + f(x) ] = a + L[/tex] (Q.E.D)

    Is there any where unclear?
    Can you do the same to part II? :)
     
    Last edited: Mar 29, 2007
  4. Mar 30, 2007 #3
    Thank you, I do have one question

    Why did you choose to divide epsilon 1 by 2?
     
  5. Mar 30, 2007 #4

    mjsd

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    because eventually you wanna sum up two of them.. and 1/2 +1/2 =1 gives you just one epsilon... and this matches the target you wanna reach. that's all
     
  6. Mar 31, 2007 #5

    VietDao29

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    Since you want to prove that:
    [tex]|[x + f(x)] - [a + L]| = |(x - a) + [f(x) - L]| < |x - a| + |f(x) - L| < \epsilon_1[/tex], i.e, the sum of |x - a|, and |f(x) - L|, so it's natural to choose the sum of [tex]\frac{\epsilon_1}{2}[/tex], and [tex]\frac{\epsilon_1}{2}[/tex], as [tex]\frac{\epsilon_1}{2} + \frac{\epsilon_1}{2} = \epsilon_1[/tex].
    Then assign [tex]\delta, \ \delta_1 , \ \epsilon[/tex] wisely, so that: [tex]|x - a| < \frac{\epsilon_1}{2}[/tex], and [tex]|f(x) - L| < \frac{\epsilon_1}{2}[/tex]
     
  7. Feb 8, 2008 #6
    I think I understand the process shown for the above proof but if someone could provide the second part it would improve my understanding. Given that lim f(x)=L as x approaches a , prove that lim x*f(x)=aL as x approaches a using the delta epsilon definition of a limit. This would be much appreciated. Thanks.
     
  8. Feb 8, 2008 #7
    just try it yourselves! Use the method and I'm confident it will work out!
     
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