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Delta-epsilon limit of a rational function with a jump discontinuity

  1. Apr 1, 2010 #1
    Hello, I am trying to prove the following...

    lim (x+3) [tex]\left|x+5\right|[/tex][tex]/x+5[/tex]

    from the left, L=+2
    from the right, L=-2

    I used delta-epsilon on the right hand limit and got [tex]\delta[/tex] = [tex]\epsilon[/tex]

    However, I'm not sure how to proceed when I get to this step while trying to prove the left hand limit:

    [tex]\left|x+1\right|[/tex]< [tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex]<[tex]\delta[/tex]
  2. jcsd
  3. Apr 1, 2010 #2


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    If you are approaching from the left then |x+5|/(x+5)=(-1). So your expression f(x) becomes (-1)*(x+3). The quantity you want to prove less than epsilon is |L-f(x)|. I make that to be |2-(-1)(x+3)|=|x+5|.
  4. Apr 1, 2010 #3
    I think I see where you're going, but I want to make sure I have the right idea with the delta-epsilon definition.

    lim f(x)=L

    [tex]\left|f(x)-L\right|[/tex] <[tex]\epsilon[/tex] if 0<[tex]\left|x-a\right|[/tex]< [tex]\delta[/tex]

    Is this the same definition that is used to find the one sided limits? If so, how is it manipulated to arrive at the conclusion you provided?

    What I was trying to do as I attempted to prove the limit from the right:

    [tex]\left|f(x)+2\right|[/tex] < [tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex] < [tex]\delta[/tex]

    Which gave me:

    [tex]\left|x+5\right|[/tex] <[tex]\epsilon[/tex] if 0<[tex]\left|x+5\right|[/tex] < [tex]\delta[/tex]

    and I called it a day on that one. But when I got to the second one, I tried using the same definition to try to prove the limit. I have a feeling that in the same format, it will give me an incorrect answer...where I am lacking understanding?
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