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Limit help, please!

  1. Sep 24, 2009 #1
    Okay, so my alarm clock magically didn't go off this morning and I missed calculus (my first class missed of the year, grr!) :(

    Unfortunately the teacher went over limits today and we have work due this weekend online with limits and I don't have class tomorrow. I'm having trouble figuring out how to solve these problems. Can someone help?


    1. The problem statement, all variables and given/known data
    The first one is really confusing, it is just a graph and it says Use the figure below to give an approximate value for the limit
    lim f(x) (if it exists).
    x[tex]\rightarrow[/tex]1
    http://img121.imageshack.us/img121/1862/picture4r.png [Broken]

    There is another problem just like this one.

    Then there is another graph problem but it is different, now I'm really confused. It says to estimate the limits using the graphs.

    http://img4.imageshack.us/img4/4196/picture5ju.png [Broken]

    I'm supposed to find:
    lim ( f (x) + g (x))=
    x[tex]\rightarrow[/tex]5^-
    lim ( f (x) + 8g(x))=
    x[tex]\rightarrow[/tex]5^+

    Another thing I don't get is the x[tex]\rightarrow[/tex]5's have + and -. What in the Universe does that mean?

    The last problem is yet again different. No graph for this one. It just says to find a value of the constant k such that the limit exists for the given function.

    http://img156.imageshack.us/img156/7294/picture6rn.png [Broken]
    http://img156.imageshack.us/img156/6981/picture7ki.png [Broken]
    2. Relevant equations



    3. The attempt at a solution

    I know this may seem like a lot to ask, but I really need help to understand this, and I'm not just some lazy person who missed a class because I didn't feel like getting out of bed. I have literally been studying physics most of the week and getting to bed late. Thanks ahead of time to anyone who is willing to help explain how this works.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 24, 2009 #2
    Honestly, it would be better if you at least skimmed the chapter on limits in your book and then asked one question at a time. It's a lot of LaTex to type up if you know what I mean.

    I'll be happy to help with any specific questions you have of course!

    The main idea with limits is that [tex]\lim_{x \to a}f(x)[/tex] is the value that "f gets closer and closer" to as "x gets closer and closer to a" if that helps you with the first ones.
     
  4. Sep 24, 2009 #3
    I have skimmed through it, it does a really bad job of explaining how to do these. I'm trying to do the first problem right now. I'm guessing you have to do something with all of the open and closed circles to figure out the limit, but I'm clueless to how to go about setting that up. :(
     
  5. Sep 24, 2009 #4
    Well if f is continuous at a then [tex]\lim_{x \to a}f(x) = f(a)[/tex] if that helps. You can just do the first few by looking at them.
     
  6. Sep 24, 2009 #5
    I understand but I'm supposed to find the value, not if they are continuous. What am I not seeing?
     
  7. Sep 24, 2009 #6
    Well, if you can see that it's continuous then you can just look and see what it's value is at 1! That's the limit.
     
  8. Sep 24, 2009 #7
    So, what you're saying is for [tex]
    \lim_{x \to a}f(x) = f(a)
    [/tex] the approximate value either has to be a or it is not continuous right?
     
  9. Sep 24, 2009 #8
    If you are looking at a graph, then yes because you can't have a precise value. But in general, when you are working with functions written as formulas, the value will be precise.
     
  10. Sep 24, 2009 #9
    I got that one wrong, I get 2 chances on every one. The answer wasn't 1 so I put na and it was wrong too. :(
     
  11. Sep 24, 2009 #10
    the answer is 8! That's the value of f(x) at 1 ;)

    Maybe you should read the chapter a little more thoroughly
     
  12. Sep 24, 2009 #11
    Oh my goodness, I'm even more confused now!
     
  13. Sep 24, 2009 #12
    Well, what's the value of f(x) at 1 in the first graph?
     
  14. Sep 24, 2009 #13
    Why not 3? And how would that change if the open circle at 8 wasn't above 1 on the x-axis?
     
  15. Sep 24, 2009 #14
    oh my bad man! It's been too long since I looked at a graph like that. Maybe I shouldn't be helping out but I think that's the only thing that's gonna trip me up. Now I remember how to read those graphs. I guess we're both gonna learn something. At least you already got the problem wrong and it wasn't my fault :P

    f(x) isn't continuous at 1 because the the value 'jumps' to 3. What is the definition of a limit that they give in the book. Maybe we should work from that.

    So the limit is 8. The intuitive idea of a limit is that it's the value that f(x) get's closer and closer too.
     
  16. Sep 24, 2009 #15
    http://img121.imageshack.us/img121/5796/picture8u.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  17. Sep 24, 2009 #16
    Oh wow! surprise! That's a real definition. They almost never do that in calc 1.

    Do you understand, given the intuitive definition that I gave, why the limit is 8 though? We should cover that first. Do you see how the graph of f(x) gets closer and closer to 8 as x gets closer and closer to 1? It doesn't actually matter what the value of f(x) at 1 is or if it's even defined!
     
  18. Sep 24, 2009 #17
    Not really, the graph is leaving me absolutely confused. I see there is an open circle at 8 when x is 1 and a closed circle at 3.
     
  19. Sep 24, 2009 #18
    That means that the value of f(x) at 1 is 3. The open circle means that the graph breaks there. An open circle is what they put on a graph to indicate that it looks like the value of f is where the open circle is but that it's actually somewhere else, or undefined. If you see an open circle, then if there is a solid circle somewhere else on the some vertical line, then that is the value of the function. So there is an open circle at 8 and a solid circle at 3.

    I had forgotten that and so thought that the function was continuous there. At least you learned the definition of continuity though.
     
  20. Sep 24, 2009 #19
    What if there is a closed circle on top and an open one at the bottom?
     
  21. Sep 24, 2009 #20
    Then the value is wherever the closed circle is. The open circle will always be on the graph though.
     
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