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Homework Help: Basic question about limits

  1. Jun 5, 2010 #1
    Hi

    This is more of an intuitive question than a question that has a definitive answer.

    When you are evaluating limits, say a limit of f(x) as x approaches 0, you can cross cancel terms of x that appear in the f(x) because x is not actually zero. However when you are done cancelling and have your limit in its simplest form it then appears as if you treat x as if it were 0 to get the limit.

    This can seem a touch confusing. In other words, 'one minute x isn't 0 but next minute we may as well assume it is.' I was wondering if any of you seasoned experts in this subject had any time-tested pearls of wisdom that make this seem less confusing.

    thanks
     
  2. jcsd
  3. Jun 5, 2010 #2
    Are you doing this because you assume the function is continuous? ( so you can plug in the value 0 for x once you have done your factorising/cancelling)
     
  4. Jun 5, 2010 #3
    When there is something small in the numerator, the fraction is small. When there is something small in the denominator, the fraction is large. If there is something small in both, it's not clear which effect dominates, so we try to simplify it into a form in which things become clear.

    EDIT: In response to your second question: if my original expression had x in the denominator, it is not continuous at x = 0 since it is undefined there. If I simplify it, I may get the same function, except it is now defined at 0. If this function is also continuous, we can plug in x = 0 to find the limit in this new, simplified function, let's call it L. But since it is the same as the original except at x = 0, we see that the limit of the original must be L as well.
     
  5. Jun 5, 2010 #4
    Hi, I think your EDIT is what I was getting at. Because x is in the denominator and we are taking the limit as x->0 you need to get the function into a form (L) where it is defined at x. Because the function is continuous at x you can plug in x=0. The limit of L and the original function are the same because they only differ at x=0 and the limit as x->0 is independent of what actually happens at f(0)

    But the key point is you can only do this if the function is continuous? Is that correct?
     
  6. Jun 5, 2010 #5
    Since the definition of continuity at a point is that the limit is equal to the value, I believe so. As long as your function is continuous at the point toward which x is tending.
     
  7. Jun 5, 2010 #6
    It sounds like this discussion is getting a little circular.

    Let's go back to basics. A function of a single variable f(x) has a limit L as x goes to a if for any [itex] \epsilon > 0 [/itex], there exists a [itex] \delta > 0 [/itex] such that if [itex] 0 < | x - a | < \delta [/itex], [itex] | f(x) - L | < \epsilon [/itex].

    So to say that something has a limit is to say that it satisfies this definition.

    Now consider f(x) = mx + b. You can easily prove that f satisfies this definition for any real m and b, and that:

    [tex] \lim_{x \rightarrow a} f(x) = f(a) [/tex]

    Now this isn't "because f is continuous," although an intuitive sense of continuity will help you later. It's because it satisfies the definition of a limit. Continuity follows from the definition a limit, not the other way around. You can prove this type of thing for a lot of different functions, like quadratics, some trigonometric functions, etc. By doing so, you are showing that they are continuous.

    Now, when you are considering limits of the type you mention, you ask about dividing out x in the numerator and denominator. This is ok because in a limit x is never actually zero; it is just arbitrarily small. But since it is arbitrarily small in both the numerator and denominator, it divides out normally and doesn't cause undefined behavior (because there is no division by zero).

    It may happen that after you have divided out something like this, the form of the resultant expression is of the form you derived above where the limit as x goes to a is the value of the function at a. But that wasn't true before you divided out.
     
  8. Jun 5, 2010 #7

    Hurkyl

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    Some limits are very easy to compute -- e.g. the limit of a continuous function. (You compute by plugging in)


    This provides the basis for one approach to computing limits: find a way to express the limit you want in terms of easy-to-compute limits.


    For example, to compute
    [tex]\lim_{x \to 1} \frac{x^2 - 1}{x-1}[/tex]​
    we first observe that, on the set of points [itex]x \neq 1[/itex]:
    [tex]\frac{x^2 - 1}{x-1} = x+1[/tex]​
    (of course, this is not an identity -- the relation fails at x=1)

    If two functions are everywhere equal on the set [itex]x \neq 1[/itex], they have the same limit at 1:
    [tex]\lim_{x \to 1} \frac{x^2 - 1}{x-1} = \lim_{x \to 1} x+1.[/tex]​
    The latter is an easy-to-compute limit; once we have computed it, we have found the original limit.


    We can get away with less than having equality everywhere except 1 -- we only need equality on a punctured neighborhood of 1. For example, to compute
    [tex]\lim_{x \to 1} |x|[/tex]​
    we note that, on the sets [itex](0, 1)[/itex] and [itex](1,2)[/itex], we have the equality
    [tex]|x| = x[/tex]​
    (actually, we have equality on all if [itex](0,2)[/itex], of course. But I wanted to emphasize the value at 1 is irrelevant) and therefore
    [tex]\lim_{x \to 1} |x| = \lim_{x \to 1} x.[/tex]​
    (Of course, we can just plug in 1 to begin with, since |.| is continuous. But this trick is useful for more complex examples)
     
  9. Jun 6, 2010 #8
    @hgfalling. Thank-you for your reply. However I think I'm now more confused than I was previously! :)

    If I am trying to evaluate the limit of a continuous function I can factorise the function and cancel terms so that I don't end up with any division by 0. At this point I can plug in the value of f(a) (if we are talking x->a) because, or so i thought, the function is a continuous function. the definition of a continous function is that its limit has the property lim(x->a) f(x) = f(a). So the limit does come first as you say as the definition of the continous function is based on obeying this limit property.

    But you can have other functions that have a limit and that aren't continuous (such as a piecewise function with the variable a defined as having a different output than everything else). With these functions you can't just plug in the value of f(a) to get the limit. So the function does have to be continous doesn't it to evaluate the limit in this way?
     
  10. Jun 6, 2010 #9
    Why wasn't that true before you divided out?
     
  11. Jun 6, 2010 #10
    Only because the function may not be defined before (i.e. division by 0).

    I think what hgfalling is point out is that formally you use limits to prove that a function is continuous. Of course, if you already know that your function is continuous from previous knowledge or proofs, and you're allowed to use that knowledge, you can plug it in. You find this a lot in high school questions where they just ask you to find the limit without proving it. When you are asked to prove something is a limit, however, you should resort back to ε-δ.
     
  12. Jun 6, 2010 #11

    Mark44

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    Can you give us an example?
     
  13. Jun 6, 2010 #12

    vela

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    You're asking does f(x) have to be continuous at x=a for the equality

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

    to hold. Well, yeah. By definition, that's what it means for f(x) to be continuous at x=a.
     
  14. Jun 6, 2010 #13

    vela

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    Let me summarize what people have said here so far. Assume the limits exist and the functions behave nicely as is typical in this type of situation.

    You start with a function f(x) that's not defined at x=a because of a division by zero. After you simplify it, you have a second function g(x) that is defined at x=a because the simplification eliminated the division by zero. Note that at points other than x=a, you have f(x)=g(x). Because a limit only depends on how a function behaves near x=a and not at x=a, f(x) and g(x) have the same limit as x approaches a:

    [tex]\lim_{x \to a} f(x) = \lim_{x \to a} g(x)[/tex]

    In addition, because g(x) is defined at x=a, it's continuous at x=a (by assumption), so you also have

    [tex]\lim_{x \to a} g(x) = g(a)[/tex]

    Therefore, it follows that

    [tex]\lim_{x \to a} f(x) = g(a)[/tex]

    You never set x=a in f(x); you only set x=a in g(x). So there's never any division by zero.
     
    Last edited: Jun 6, 2010
  15. Jun 6, 2010 #14
    Vela, thanks for your reply. Its very clear. However, I don't think you are addressing the question I am asking, probably because I'm not asking it very clearly. I think Tedjn sees what I am getting at.

    Maybe I should try again...

    When you evaluate your limit in the end step by setting x=a in g(x) (and not f(x) as you explaiend very clearly) you can substitute x=a because you know your function is continuous (I'm assuming you know this from one thing or another). However you could have started with a piecewise function where the function is not continuous at a because the value at a is set to be something different than every other input value. YOu couldn't just set x=a there could you? I've not come across an example like this in the book I'm reading. I was just wondering.
     
  16. Jun 6, 2010 #15

    vela

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    You're right. You can't just blindly plug in x=a in that case. You need g(x) to be continuous, and I assumed that after you simplify f(x), the resulting g(x) is, in fact, continuous, like in the example Hurkyl gave above. You could always define weird functions where g(x) isn't continuous at x=a, but then you don't have that

    [tex]\lim_{x \to a} g(x) = g(a)[/tex]

    so you can't conclude that

    [tex]\lim_{x \to a} f(x) = g(a)[/tex]

    I think that answers your question. If not, as Mark suggested, try providing us with an example to illustrate what you mean.
     
  17. Jun 6, 2010 #16
    That does answer my question. I don't have an example. I was just wondering what would happen if that situation arose. And i learnt a lot from asking so thank you all for your help
     
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