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The definition of a limit.

  1. Apr 10, 2012 #1
    Hello all,
    This is very simple however I would like to understand why this is true.

    According to the definition of a limit, if we have limit of f(x) as x approaches infinity = a
    then for every ε>0 there exists a real number M such that if x>M then the absolute value of f(x)-a < ε.

    This is very strange to me. I did not study this definition when studying limits before. I know what limits mean but this definition seems odd to me. If someone could explain why this is true I would greatly appreciate it. Also, is the definition the same when the limit approaches any x other than infinity?

    Thanks in advance.
     
  2. jcsd
  3. Apr 10, 2012 #2

    chiro

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    Hey MinaHany and welcome to the forums.

    Basically you can think about this in the way that there is some point for x where after this x-value, the absolute difference difference between the function at x = a (i.e. f(a)) is less than this epsilon. Also I think you will have to modify this if you are considering a proper limit which is two-sided.

    Now to think about when this is not true, consider say a discontinuous function like the Heaviside function. If we take the limit for a = 0 from the left we get our limit to be 0 but if we take it right from the right, then we get 1. Since 1 != 0 the limit in this case doesn't exist.

    With our ε values we would show that we would get contradictions if we compared the left hand side with the right if we tried for example to show that the limit was either 1 or 0 because we would get a contradiction in both cases since LHS limit suggest 0 but RHS suggests 1 and with our ε we can show that for a small ε of say 0.1 for LHS we will get a contradiction for RHS limit and vice-versa.

    Now in your case the reason why we only look at one side is that the limit has to do with +infinity which is why it's one sided, but I feel it's important to point out that limits are usually considered in a two-sided case and this is a unique case because we are dealing with an infinity in one dimension.

    For the infinite-dimensional case if you want an example of why this might fail, consider the function f(x) = sin(x). Try and choose any ε > 0 and you'll always find a contradiction because f(x) does not converge at infinity at all: it just keeps oscillating up and down and therefore you can not define f(x) at x = infinity.

    If you are stuck, try a specific value of a and show that for a really small ε > 0 there is always a value of |f(x) - a| that is larger for some x > M. When you do this you have proven a contradiction since it has to hold for all ε > 0 and not just one particular case.

    Once you get used to this you'll be able to just introduce some mathematical tricks with variables to prove things generally.
     
  4. Apr 10, 2012 #3
    Thanks a lot for your fast reply chiro.

    I understood most of what you said perfectly. I know when a limit does not exist and the contradiction between LHS and RHS, etc..

    There's only one part I didn't quite understand:
    "Basically you can think about this in the way that there is some point for x where after this x-value, the absolute difference difference between the function at x = a (i.e. f(a)) is less than this epsilon."

    what do you mean the absolute difference between the function at x=a?
     
  5. Apr 10, 2012 #4

    chiro

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    Basically i am referring to |f(x) - a| < ε. Also because this limit is one-sided (you can't approach +infinity from the right hand side in normal geometry) this is why you have x > M instead of something else, but to answer your question see the expression in my first line of this reply.
     
  6. Apr 10, 2012 #5
    I think I'm close to understanding the whole thing. Would you please explain this with a simple example?
     
  7. Apr 10, 2012 #6

    chiro

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    Well you could use two examples for which in one the limit exists and in the other it doesn't.

    The first one would be y = e^(-x) and the second one would be y = sin(x). Both examples are going to be based around showing the limit exists for x -> +infinity.

    I just want to point out that I'm not really good at this, but I'll so how we go.

    First consider a function that has no limit: f(x) = sin(x)

    Let f(x) = sin(x).

    Now suppose our limit is a for some a.

    now |f(x) - a| < ε. Let's choose ε = 1/8 for x = M. Now consider x = M + pi/2. Now sin(M + pi/2) = sin(M)cos(pi/2) + cos(M)sin(pi/2) = cos(M). Now a = sin(M) which means |sin(M) - a| < 1/8 for this ε for all x > M.

    Using all of this we can show that if |sin(x) - a| < ε, then |cos(x) - a| < ε for all x > M for some real number M. Squaring both both sides (since both LHS and RHS of each equation is positive) gives us:

    sin^2(x) - 2sin(x)a + a^2 < ε and cos^2(x) - 2cos(x)a + a^2 < ε. This means that since both are positive we get:

    sin^2(x) - 2sin(x)a + a^2 + cos^2(x) - 2cos(x)a + a^2 < 2ε for all ε > 0. But for the LHS we get:

    (sin^2(x) + cos^2(x) + 2a^2 - 2a[sin(x) + cos(x)] < 2ε
    1 + 2a^2 - 2a[sin(x) + cos(x)] < 2ε for all ε > 0 and all x > M.
    1 + 2a^2 - 2a[sin(x) + sin(x+pi/2)] < 2ε for all ε > 0 and all x > M.

    Let x = 2npi - pi/4 for all integer value of n where x > M. This means that for this x, sin(x) + cos(x) = 0 which implies:

    1 + 2a^2 - 2a x 0 < 2ε for these particular values of x which implies
    1 + 2a^2 < 2ε

    But this is a contradiction since it has to hold for all ε > 0 and this is not the case for any value of finite a both positive and negative. Thus f(x) = sin(x) has no limit as x -> +infinity.

    Would you like me to do one where you have a limit or is this ok?
     
  8. Apr 11, 2012 #7
    Wow! Thanks a lot for this it was very helpful.
    Would I be asking too much for an example where the limit exists? Just a very simple one will do.
    Sorry for all the questions chirno!
     
  9. Apr 11, 2012 #8

    chiro

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    Basically something like f(x) = e^(-x) or f(x) = 1/(1 + x). You can think of the limit existing where the further you go, the difference between the limit and the function gets small enough as to approach zero.

    Note that the limit does not have to equal the function value. This is why you have the epsilon greater than zero but not actually zero itself.

    For all we know the function at infinity could have a value of infinity or a + 2 where a is the actual limit. But the reason limits are useful is that if you can show the limit equals the value of the function for every part of the domain, then the function is continuous and if the function is continuous then it makes it a lot easier to analyze than if it wasn't.
     
  10. Apr 11, 2012 #9
    Thanks a lot for all your help chiro! That was very useful.

    Have a nice day :)
     
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