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Interpreting a function based on it's equation.

  1. Feb 11, 2012 #1
    Ok so firstly, I'm not entirely sure where this thread should land so it's going in here lol.

    So this question is inspired by doing preliminary algebra on a function in order to find a limit that initially results in an indeterminate form.

    What does it say about what we get from the face value of a written equation, and what it actually is? For example,

    lim x→1 of (x^2 - 1) / (x - 1)

    makes the exact same looking graph on a calculator as (x-1) , but the first equation does not exist at 1, whereas the second one does! BLLAAAARRGGHHHHH!!!!! o_O
  2. jcsd
  3. Feb 11, 2012 #2


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    Hey Square1 and welcome to the forums.

    For this kind of problem you are indeed right in the confusion that arises and this is why there is a lot of 'funny things' that happen when you get the 0/0 form.

    Having said that, there are rigorous proofs that allow us to evaluate limits such as these in certain ways and one of them is L'Hopitals rule which allows you figure out these kind of situations.

    Also what I am about to say is the most important fact of them all: Limits are used to figure out the behaviour of something when it gets infinitesimally close to a value but not at the value!

    It seems crazy, but this is what calculus is based on. So by doing this we are not really considering a 0/0 case but a case where it gets really close to zero but not actually zero!

    It is for this reason that we can do things like factor out terms since (x^2 - 1)/(x-1) = (x-1)(x+1)/(x-1) = (x+1) as long as x does not equal 1. But the limit doesn't talk about x actually equalling 1, but instead when it gets really really close to 1 (but still not actually 1!)

    This is why we can do this kind of thing because we are not actually talking about evaluating a 0/0 but figuring out what it goes to as we approach it.

    Then what you do is you take this result and add what is called continuity and what this will do is help you specify that the limit as well as the function value corresponding to information contained in the limit are the same. In other words if the derivative is analytic-continuous and the function is continuous then the information given by the limit represents the actual function itself.

    I can go into this in more detail and clarify these things if you want, but the above should give you the idea of why we can do what we do.
  4. Feb 11, 2012 #3


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    The graphs of y = (x2 - 1)/(x - 1) and y = x + 1, except at x = 1, where the first function has a discontinuity at the point (1, 2). A single point has no width or length, so a graphing calculator won't show this discontinuity.

    Note that you said that the first function has the same graph as y = x - 1, which isn't true.
  5. Feb 12, 2012 #4
    Well, the calculator failed basic algebra! Because dividing both the numerator and the denominator by x - 1 is only permissible when x doesn't equal 1. When x = 1 we can't do the division; and the function is not defined there.
  6. Feb 12, 2012 #5


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    The graph of x+ 1 is a straight line. The graph of [itex](x^2-1)/(x- 1)[/itex] is that same straight line except that it has a "hole" at (1, 2).
    Last edited by a moderator: Feb 12, 2012
  7. Feb 12, 2012 #6
    Ok good point and you are right.

    Also, yea as the second post says, its good to keep in mind the distinction between approaching a point and actually being on the point.

    But hmm. How to put it...It's just kind of hitting me that the single process of simplification has the inherent property of taking an equation, and creating another one that is able to "pass the test" so to speak of finding that limit, and the answer applies to the first equation!

    Ugh I feel as if im not explaining this right. I'll just wait and see
  8. Feb 12, 2012 #7
    You're right. It's a little counterintuitive to realize that f(x) = x is the not same function as f(x) = x^2/x. But as I indicated above, the equality x^2/x = x is only valid if x is different from 0. So f(x) = x^2/x is not defined at zero; although of course the limit as x->0 of x^2/x is in fact 0.
  9. Feb 12, 2012 #8


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    the situation is a little clearer if one considers the function:

    f(x) = x/x

    (now we don't have any complicated algebra in the way).

    well, as any fool can tell you, x/x = 1, right?

    well, almost. 0/0 doesn't make any sense, because __ /0 is "bad" (undefined).

    but, f(x) = x/x is perfectly reasonable everywhere EXCEPT 0, so if we made THIS function:

    g(x) = x/x, x ≠ 0
    g(0) = 1

    the function g behaves the way we WANT f to behave.

    in technical terms, g is CONTINUOUS, while f is NOT.

    continuous functions, therefore, are functions that "make the sense they're supposed to". non-continuous functions, might do something "unexpected", so it might be best if we save studying those for later, after we understand continuous functions better.

    of course, what i've done with the above discussion, is said basically:

    if a function is continuous, certain kinds of troublesome problems go away.

    well, that's all very well and good, but what, exactly, IS a continuous function (we've traded difficulty with functions, off for difficulty with some definition of a property a function might have)? and that's where the concept of a LIMIT comes in.

    a limit acts like a number, or rather, acts like an ordinary "output" of a function, but it "extends the places a function can go". roughly speaking, a continuous function "equals its limits at all points".

    now, we've made yet ANOTHER trade-off...to have any hope of knowing what continuous means, we need to know what a limit is. naively, a limit of a function, is what the output is near, when the input is near a given spot, say x = a. the exact definition of "near" requires strange quantifiers with epsilons and deltas, but in a loose sense, x near a means that the difference between x and a is small (close to 0, in a well-defined way).

    so a continuous function is one for which f(x) is near f(a), when x is near a. this corresponds nicely to our naive idea that a continuos function of one variable, is something you can draw "without the pencil leaving the paper".

    as you can see with f(x) = x/x, when we graph it, we have a microscopic gap at x = 0. such a small hole won't show up very well on a graph, but is disastrous if f represents something like an internet connection (the connection gets "broken" at x = 0, and that's that).

    this is something to watch out for, for any function of the form:

    f(x) = g(x)/h(x).

    we might have h(a) = 0, for some value a, which might make f do some strange stuff near a.
  10. Feb 18, 2012 #9
    Ok I'd like to have another go at this so I'll rephrase the question.

    How is it that the process of factoring, expanding, or multiplying by a conjugate produces almost an identical function as the original.

    A)What is so special about these processes that they can reproduce ALMOST THE EXACT SAME THING, ONLY WITH THE EXCEPTION OF A FEW POINTS????

    B) so.....(x^2-4)/(x+2) != (x-2) :O ??
  11. Feb 18, 2012 #10


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    When you factor an expression, all you are doing is writing that expression in a new form that is identically equal to the original form. For example, x2 + 2x + 1 = (x + 1)2 is identically true (i.e., true for all values of the variable x).

    When you expand an expression, you are just multiplying factors to get another expression that is identically equal to the first. In my equation above, you can expand (x + 1)2 to get the expression shown on the left side.

    You can always multiply an expression by 1 to get a new expression that has exactly the same value. If you multiply an expression by the conjugate, you will get a new expression with a different value. However, if you multiply the expression by the conjugate over itself, you are just multiplying by 1, so the new expression is identically equal to the one you started with.
    Here we are factoring the left side to get
    [tex]\frac{x^2 - 4}{x + 2} = \frac{(x-2)(x + 2)}{x + 2)} = (x - 2) \cdot \frac{x + 2}{x + 2}[/tex]

    The fraction on the right is equal to 1 except when x = -2, which also happens to be where the original rational expression on the left side is undefined.
  12. Feb 18, 2012 #11
    Hmm I think I am coming closer. But still some q's..

    Hmm so both right and left side continue to be undefined at -2, and yet in one expression it is possible to calculate the value at -2!

    I see the idea behind saying that (x+2) times (x-2)/(x-2) is like saying times 1, but (x-2)/(x-2) is only equal to one if x!=0. It is indeterminate form right?
  13. Feb 18, 2012 #12


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    Its only equal to 1 when x != 2 since (x - 2) = 2 -2 = 0 which gives an indeterminant form expression.
  14. Feb 18, 2012 #13
    Yea so thats kind of what I had said, and to repeat, we still end up calling (x-2)/(x-2) = 1 when we plug in -2, even though (x+2) times (x-2)/(x-2) is still, as said, undefined at -2.
  15. Feb 18, 2012 #14


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    This kind of thing is actually why we have to use limits in calculus.

    In calculus we use limits to define things like the derivative and the reason we don't just plugin the 'h = 0' in place of 'h approaches 0' is for a similar reason in that we get an indeterminant form. We can't just let h = 0 because the same kind of situation that you have brought up.

    But this kind of thing is a weird thing and you're definitely not the first person to ask about it.
  16. Feb 19, 2012 #15
    haha yea. it definatly is occupying my thoughts at the moment :)
  17. Feb 19, 2012 #16


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    Not in what I wrote above. The three expressions in my equation are all undefined at x = -2.
    You have switched some signs, and have another error. (x - 2) times (x + 2)/(x + 2) is the same as x - 2 times 1 if x != -2. The expression (x + 2)/(x + 2) has the indeterminate form [0/0].
  18. Feb 20, 2012 #17
    The limit of the rational function is indeterminate because the limit of the functionn depends on whether the numerator approaches zero or the denominator approaches zero.
  19. Feb 20, 2012 #18


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    If the rational function you refer to is (x2 - 1)/(x - 1), the limit as x approaches 1 is NOT indeterminate.
    [tex]\lim_{x \to 1}\frac{x^2 - 1}{x - 1} = 2[/tex]
  20. Feb 20, 2012 #19
    Uh....the form is indeterminate. That is why you can use L'Hopital's Rule to find the limit.
  21. Feb 20, 2012 #20


    Staff: Mentor

    Yes, the expression is one of several indeterminate forms, but the limit is 2, which is not indeterminate. In any case, a limit is never indeterminate.
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