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Epsilon-delta question with limits

  1. Feb 5, 2008 #1

    danago

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    Gold Member

    Let [tex]
    f(x) = \left\{ {\begin{array}{*{20}c}
    x & {x < 1} \\
    {x + 1} & {x > 1} \\
    \end{array}} \right.
    [/tex], and let [tex]\varepsilon = 0.5[/tex]. Show that no possible [tex]\delta > 0[/tex] satisfies the following condition:

    For all x:
    [tex]0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| < 0.5[/tex].

    That is, for each [tex]\delta > 0[/tex] show that there is a value of x such that:

    [tex]
    0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| \ge 0.5
    [/tex]

    This will show that [tex]\mathop {\lim }\limits_{x \to 2} f(x) \ne 2[/tex]


    I started by finding the set of x values for which the function produces values between 1.5 and 2.5.

    For x<1:
    [tex]
    \begin{array}{l}
    \left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 2} \right| < 0.5 \Rightarrow x \in (1.5,2.5) \\
    (1.5,2.5) \cap ( - \infty ,1) = \emptyset \\
    \end{array}
    [/tex]

    Therefore there are no values of x<1 for which f(x) is within 0.5 of 2.

    For x>1:
    [tex]
    \begin{array}{l}
    \left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 1} \right| < 0.5 \Rightarrow x \in (0.5,1.5) \\
    (0.5,1.5) \cap (1,\infty ) = (1,1.5) \\
    \end{array}
    [/tex]

    Therefore, f(x) is within 0.5 units of 2 if and only if [tex]x \in (1,1.5)[/tex]. This means that the only possible value for delta is 0, and since [tex]\delta > 0[/tex], this value of 0 is not possible, thus making no possible value for delta.

    Is that how i should go about proving such a limit does not exist? Are there flaws in what i have done?

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. Feb 6, 2008 #2
    are u required to use only delta epsylon to show that the lim of this function as x-->1 does not exist, because if not there is a nicer way using sequences to show this?
     
  4. Feb 6, 2008 #3
    I can show you a general way using sequences, and also incorporating delta and epsylon in it, to show that this limit does not exist. YOu might want to try to use the Bolcano-Cauchy criterion for the existence of the limit. Are u familiar with it?
     
  5. Feb 6, 2008 #4

    danago

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    Gold Member

    No i cant say i am familiar with it :( Im only just about to begin my first year in an undergraduate course in engineering, and thought id try and get a head start in the calculus.
     
  6. Feb 6, 2008 #5
    well it basically says that a function f is said to have a limit A, as x-->a if and only if for every epsylon there exists some delta such that for any two points from its domain, that is x',x"
    that satisfy the condition abs(x'-a)<delta and abs(x"-a)<delta , than the following is fullfilled:

    abs( f(x')-f(x"))<epsylon.

    so the idea to show that your function does not have a limit as x-->1 is to find any two points x', and x'' ( also sequences if you like to) and to show that

    abs(f(x')-f(x")) will always be greater than epsylon =.5

    such numbers would be x'=(1-1/n) where n is a natural number, and x"=(1+1/n)
    look these two numbers are one at the right of 1 and the other at the left. Do you see how to go now?
     
  7. Feb 6, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Obviously, if x< 1, there exist x arbitrarily close to 1, that is |x-1|< [itex]\delta[/itex], such that f(x)= x is close to 1 and |f(x)- 2| is close to 1 not < 0.5.
     
  8. Feb 6, 2008 #7
    yeah, x'=(1-1/n) is close to 1 and x'<1, so obviously

    abs(x'-1)=abs(1/n)<delta, whatever you chose delta to be but on the contrary

    abs(f(x')-2)=abs(-1) which is obviously greater than 0.5.

    But you can use the bolchano-cauchy's criterion as well, as i explained above, and you will get to the same result, depends which one you like more!
     
  9. Feb 6, 2008 #8
    Because in these cases where u are given to prove that the limit of some function is not, say A, the method above will work, because it simply contradicts the deffinition of a limit given my Cauchy. But if you are asked to prove whether the limit of some function exists or not, in general withoug given to prove whether the limit of f is A, than the method above wont be of any help. SO, in those cases in order to determine whether the limit of that particular function exists or not as x-->a, you need to use the Bolcano-Cauchy's criterion for the existence of the limit, as i explained in my previous posts.
     
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