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Delta Epsilon Proof of a Limit

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Prove, using the formal definition of limits:

    If [PLAIN]http://rogercortesi.com/eqn/tempimagedir/eqn4201.png [Broken] and c>0, then [PLAIN]http://rogercortesi.com/eqn/tempimagedir/eqn4201.png [Broken] [Broken] (add the constant beside f(x) here, I couldn't get the equation generator to cooperate)

    2. Relevant equations

    3. The attempt at a solution

    My text proved one kind of similar, so using that I get this:

    Since [PLAIN]http://rogercortesi.com/eqn/tempimagedir/eqn4201.png [Broken] [Broken] for e/|c| > 0,
    there exists a d > 0 such that
    |cf(x) + c*inf| < e/|c| for 0 < |x-inf|< d

    |cf(x) + c*inf| = |c||f(x) + inf| < e/|c|*|c| = e for 0< |x-inf|< d.

    I'm probably way out to lunch here... what do the pros think?

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 5, 2011 #2
    But you have to define f(x) for this to make sense. There are a lot of functions that does not go to infinity as x goes to infinity. For instance: f(x)=1/x.
  4. Nov 5, 2011 #3


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    to make a formal proof of this, you need to use the formal definition of what:

    [tex]\lim_{x \to \infty} f(x) = -\infty[/tex]


    namely, for every positive real number M, there exists a positive real number N, such that

    x > N implies f(x) < -M.
  5. Nov 5, 2011 #4


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    Not true.

    Jimbo57 is given that [itex]\displaystyle \lim_{x\to\,+\infty}f(x)=-\infty\,.[/itex]

    He doesn't have to assume anything else about f(x).
  6. Nov 5, 2011 #5
    Hmm, so what exactly am I missing? I'm seeing people say I have to define what the functions means and some saying I don't?
  7. Nov 5, 2011 #6


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    you have to use the formal definition of such limits to prove what you wish to prove is true.

    you do not need to specify f, but you do need to use one of its properties.
  8. Nov 5, 2011 #7
    Ahh okay, so I add:

    1. For any e>0 there is a d>0 such that for any 0<|x-inf|< d it is the case that |cf(x)-c*inf|< e.

    Does that look correct?
  9. Nov 5, 2011 #8


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    no, because infinity is NOT a number. the expression |x - ∞| < d is meaningless.

    limits involving infinity are tricky, we're using "shorthand".

    suppose we want to say:

    [tex]\lim_{x \to \infty} f(x) = L[/tex]
    what does that mean? we can never "get" to infinity, it's "too far".

    what we really mean is: given any ε > 0, there is a positive real number N such that:

    x > N means |f(x) - L| < ε.

    so let's say we want to prove [tex]\lim_{x \to \infty} 1/x = 0[/tex]
    we need to find a positive real number N such that x > N means |1/x - 0| < ε.

    well, |1/x - 0| = |1/x| = 1/|x|, and since N is positive, and x > N, x > 0, so |x| = x.

    so |1/x - 0| = 1/x. so how do we pick N? let's try N = 1/ε.

    if x > 1/ε, then 1/x < ε...aha! that will work.

    now, what do we mean when we say

    [tex]\lim_{x \to a} f(x) = -\infty[/tex]
    we mean that given any positive real number M, we can pick a δ > 0 so that when:

    0 < |x - a| < δ, f(x) < -M. we can't speak of a value f(x) being "near -∞", but we can say that f(x) is less than any other negative number (unbounded below).

    when you combine the two conditions, you get what i said in my previous post. that is the definition you have to use, if you hope to prove this with any certainty.
  10. Nov 5, 2011 #9
    Okay, that helps quite a bit. I'm really curious why the text doesn't explain half of what you mentioned? Should a student studying calculus for the first time be able to assume so much?
    The math I've done isn't tough at all but this proof stuff needs a better "beginners manual".
  11. Nov 5, 2011 #10


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    a lot of times introductory courses play fast and loose with the "infinitely big" and the "infinitely" small. this can get you into trouble, which is one of the main reasons we use limits in the first place:

    [tex]\lim_{x \to 0^+} 1/x = \infty[/tex]

    does NOT mean: 1/0 is infinity!

    what it means is: as we get close to 0 (from the right), 1/x gets very large, and by getting close enough to 0, we can make 1/x as large as we like (you pick a number M, and i can pick a (positive) number x so close to 0, that 1/x is bigger than the M you picked).

    in calculus, when you see ∞ used, you should perform a "mental translation" along the lines of "larger than any (finite) positive real number M". because what we're dealing with (in calculus), is functions of real numbers, and if we want to use ∞, even loosely, we need a way of expressing what we mean in terms of ACTUAL real numbers (and ∞ doesn't qualify, because the arithmetic is dodgy).
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