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Limits Understanding

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Assume that the limit, as x approaches 0, of f(x)+g(x) exists. Must the limit be f(0)+g(0)? If it's true, explain. If it's false, give an example.


    I am failing to understand this problem. How do I check if it's true or false?
     
    Last edited: Sep 17, 2010
  2. jcsd
  3. Sep 17, 2010 #2

    Mark44

    Staff: Mentor

    I don't understand it, either. Are you sure you have stated the problem verbatim? It seems to me that you are missing something important in the problem statement.
     
  4. Sep 17, 2010 #3

    Sorry, I forgot to include the other part of the question. Must the limit be f(0)+g(0)?
     
  5. Sep 17, 2010 #4

    HallsofIvy

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    In other words, "If [itex]\lim_{x\to 0} f(x)+ g(x)[/itex] exists, must it be equal to f(0)+ g(0)"?

    Saying that [itex]\lim_{x\to a} F(x)= F(a)[/itex] is essentially saying that F is continuous at x= a, isn't it? Is there any part of your problem, that you have not told us, that says f and g are continuous at x= 0?

    Suppose f(x)= 5 if x is not 0, f(0)= 100, g(x)= 4 if x is not 0, g(0)= 100. What is f(x)+ g(x)? What is [itex]\lim_{x\to 0} f(x)+ g(x)[/itex]? What is f(0)+ g(0)?
     
  6. Sep 17, 2010 #5

    Mark44

    Staff: Mentor

    To add to what HallsOfIvy said about continuity, consider f(x) = (x2 - 4)/(x - 2)

    It can be shown that [tex]\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4[/tex]
    so this limit exists. Does that mean f(2) exists?
     
  7. Sep 17, 2010 #6
    f(x) + g(x) would be equal to 9 while f(0) + g(0) would be equal to 200. The only thing that matters is how f(x) + g(x) is defined near 0, but never considering x = 0. Is that correct?
     
  8. Sep 17, 2010 #7
    No, f(2) is undefined; therefore, [tex]\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4[/tex] cannot be equal to f(2). The limit does not care about x = 2, it only cares about how it's defined near 2. Am I correct?
     
  9. Sep 17, 2010 #8

    Mark44

    Staff: Mentor

    Yes. Now can you answer your original problem?
     
  10. Sep 17, 2010 #9
    Yes, thank you Mark44 and HallsofIvy.
    The easiest way to prove that the original problem is false is to substitute an equation for either f(x) or g(x) where f(0) or g(0) is undefined. So, I could use [tex]\lim_{x \to 0} {f(x)} + \lim_{x \to 0} {g(x)} = \lim_{x \to 0} \frac{1}{x^2} + \lim_{x \to 0} {x} = \infty[/tex]
    Since f(0) is undefined for f(x), the original is problem is false.
    Is that correct?
     
  11. Sep 17, 2010 #10

    Mark44

    Staff: Mentor

    No, that's not a good counterexample. You want lim (x -> 0) (f(x) + g(x)) to exist, but f(0) + g(0) to be undefined. Think about what was said in this thread about continuity.
     
  12. Sep 17, 2010 #11

    Char. Limit

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    But that limit does exist, doesn't it? It approaches the same value (positive infinity) from both sides of the limit, after all...
     
  13. Sep 17, 2010 #12
    I'm fine with anything other than limits as x approaches 0. Can you give me a counterexample for the original problem that works? For limits as x approaches 0, the only thing I can think of to make f(0) or g(0) undefined is an infinity function.
     
  14. Sep 17, 2010 #13

    Mark44

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    Only in the sense that infinity exists. What I was hoping for was something where the limit was a finite number, but the two functions were undefined.
     
  15. Sep 17, 2010 #14

    Mark44

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    f(x) = x2/x has a removable discontinuity at x = 0.
     
  16. Sep 17, 2010 #15
    [tex]\lim_{x \to 0} {f(x)} + \lim_{x \to 0} {g(x)} = \lim_{x \to 0} \frac{x^3}{x} + \lim_{x \to 0} \frac{x^2}{x} = 0[/tex]

    f(x) + g(x) = 0
    f(0) + g(0) = undefined

    Is that correct?
     
  17. Sep 17, 2010 #16

    Mark44

    Staff: Mentor

    Almost. f(x) + g(x) = x2 + x, if x is not 0.
     
  18. Sep 17, 2010 #17
    How would you write that as a complete answer?


    I would also like to know if:

    [tex] \infty + a finite number = \infty[/tex]

    undefined + defined = undefined
     
  19. Sep 17, 2010 #18

    Char. Limit

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    Yes.
     
  20. Sep 17, 2010 #19
    Thank you for your help, Char. Limit, HallsofIvy, and Mark44.
     
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