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Limits. Need recheck.

  1. Sep 24, 2011 #1
    dcvg45.jpg

    We have to graph these functions where they are real. (no need to graph where complex)

    d)
    For d Im getting a point at (1,0) and then half a c shaped graph opening towards the right. Is
    this correct?
    g)
    For g I have 0 on the negative side. and a straight line in the first quadrant. :s Im not sure about this one.


    ina420.png
    I have to find the limit.
    I realize this has two different directional limits positive and negative infinity. does that mean the actual limit does not exist?

    30tnhg2.jpg

    Is this very easy? I mean iim confused do i simply have to plug in underroot etc or is it something much more complicated :S
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Half correct. What about t<1?

    No, that is not correct. You should have a "stair step". If x is between 0 and 1, n= 0 so the value of the function is 0. For x between 1 and 2, n= 1 so the value of the function is 1. If x is between 2 and 3, n= 2 so the value of the function is 2, etc.\

    Yes, any time you have a fraction where the denominator goes to 0 and the numerator does not, the limit does not exist.

    It looks to me like there is a serious typographical error in this problem! The point of the exercise is to show that, as x gets closer and closer to 2, f(x) gets closer and closer to 4 and then show that you can use the same argument with the general [itex]\epsilon[/itex] as you did with .01 and .001.

    (a) asks "What is the value of [itex]x_1[/itex] such that f(x)= 2+ 0.01". That obviously should be "f(x)= 4+ 0.01"! And (b) should be " "What is the value of [itex]x_1[/itex] such that f(x)= 4- 0.01". f(x)= 4+ 0.01 is the same as [itex]x^2= 4.01[/itex]. Find [itex]x_1[/itex] by taking the square root of both sides.
     
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