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Basic limits and continuity help

  1. Jan 17, 2008 #1
    So I'm trying to grasp the epsilon,delta definition of limits.(Well not really,I'm actually just trying to be able to get the majority of the related questions right.)
    For example:
    when taking limits of rational functions:


    A result of
    0/0 is indeterminate form(suggesting a hole in the function)
    integer/0 does not exist(or does the limit go to infinity??)
    0/integer is a decreasing function(the limit is 0??)


    are these good rules of thumb or just a lot of hogwash that I've adopted ?

    also questions like these:

    use the intermediate value theorem to show that there is a solution of the given equation on the interval [1,2]

    2x^3 - 4x^2 +5x -4 = 0.

    unfortunately there are several types of questions in my book which I am required to know how to answer,yet the book provides insufficient(in my brain)explanation.(Basically I'm just asking not to flame me coz I cant attempt that question...)

    A continuity question that I'd like checked because I think there's an error in my books provided answer:
    State the types of discontinuities present for the piecewise defined function:
    h(x) 2x+9, when x<2
    x^2+1 when -2<x<=1
    3x-1, when 1<x<3
    x+6, when 3<x

    My book states that there is a removable discontinuity at x=-2...

    Your help will be greatly appreciated!!
     
  2. jcsd
  3. Jan 17, 2008 #2

    HallsofIvy

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    No, that is not exactly correct. You might still find the limit does not exist. Of course, if the limitl exists, then there is a "hole in the (graph of) the function".

    Assuming that "integer" is not 0, the limit is either infinity or negative infinity.

    Again, you are assuming "integer" is not 0. No, the function is NOT necessairily decreasing. For example (1/(x-1)/(1- x) as x goes to 1 is of the form "0/ integer" but is an increasing function. Of course, the limit is still 0. That part is correct. That follows directly from the "law of limits": if f(x)-> A and g(x)-> B, with B not equal to 0, then f(x)/g(x)-> A/B. 0 over any non-zero integer is 0.

    They are not "rules of thumb", they are both easily provable theorems that are given in any text on limits.

    The "intermediate value theorem" says: If f(x) is continuous on the interval [a, b], then f takes on every value between f(a) and f(b), for some x in [a, b]. Is 2x3- 4x2+ 5x- 4 continuous on [1, 2]? What are f(1) and f(2)? Is 0 between f(1) and f(2)?

    I am very glad you included that "in my brain". If you decide it is entirely the books fault, you have given up.

    (I have corrected "when x< 2" to "when x< -2" above.)

    Definition of "removable discontinuity": A function, f, has a "removable discontinuity" at x= a if and only if the [itex]\lim_{x\rightarrow a} f(x)[/itex] exists but f either is not defined at a or f(a) is not equal to that limit. The term "removable" is, of course, because you can "remove" that discontinuity just by redefining f(a) to be that limit.

    Since h is defined "piecewise" here, the best way to find the limit (if it exists) is to find the "one-sided limits". If x< -2, then h(x)= 2x+ 9. That's a simple linear function so it is continuous. Its limit, as x goes to -2 is the value 2(-2)+ 9= -4+ 9= 5. If -2< x<= 1, then h(x)= x2+ 1. Again, that is continuous so the limit at -2 is just the value, (-2)^2+ 1= 4+ 1= 5. Since those two are the same, the limit itself is 5. However, the value, h(-2) is NOT defined. Yes, that is a removable continuity. Defining h(-2)= 5, which we could do by including "=" on either "x< -2" or "-2< x<=1", would make that a continuous function.

     
    Last edited: Jan 17, 2008
  4. Jan 17, 2008 #3
    Apologies for the typing error,that could have been disastrous!
    I understand that 0/0 does not guarantee a hole in the function,it merely suggests it.But I have not yet figured out how to confirm the limit/hole.For example:
    if I have x-2/x-2 as x->2...I would end up with 0/0.Now that i suspect that there is a point discontinuity,how do I confirm it,by plugging in values close to 2 for both of 2's one-sided limits? ie. x->2 from the negative side --I would plug in 1.99999?
    ie. x->2 from the positive side--I would plug in 2.000001?

    When I mention "integer/0" I was referring to the result after I had substituted into the function...

    ie. (x+2)/(x-2) as x->2
    Result would be equal to 4/0.So as you stated, infinity or negative infinity would be the case here.But at this point I've exhausted my methods of finding limits..(substitute,factorise) would I have to plug in values close to 2 determine if it is negative or positive infinity??

    In my calculus book in cases like these it states "limit does not exist"...that is equivalent to saying +/- infinity??I do understand what removable discontinuities are..they are the "holes" where the function is undefined,yet a limit still exists.

    Onto the intermediate-value theorem:
    In layman's terms:Does it not mean that if f(1) and f(6) are the endpoints ,then f(anything between 1 and 6) must have a value between that of f(1) and f(6)?
    So for: 2x^3- 4x^2+ 5x- 4
    f(1) = -1
    f(2) = -2
    So the above function takes on values between [-1,-2] for any x between 1 and 2.
    So I have just solved for f(1) and f(2)...I would assume that the function is continuous on the whole interval as it is a simple polynomial with no square roots....and f(x) would be = f(c) because I am choosing my value of f(c) each time.No,0 would not be on the interval as the range is [-1,-2] (although it sounds alien to me to assume that because f(x) is continuous on some interval [1,2] that x-values between [1,2] would result in y-values between f(1) and f(2)).Is that not what the theorem states??
    I'm still lost on this one.

    Your explanations are very clear and concise,thank you
     
    Last edited: Jan 17, 2008
  5. Jan 17, 2008 #4

    HallsofIvy

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    You shouldn't have to do that. As long as x is not equal to 2, (x-2)/(x-2)= 1. The limit is 1 and (x-2)/(x-2) will be close to 1 as long as x is close to 2 whether larger or less than 1.


    Yes, but 0 is still an integer, so that could include 0/0. That was my point!

    For x close to 2, on either side, x+ 2 is close to 4. If x< 2, then, subtracting 2 from both sides, x- 2< 0: x-2 is a small negative number. 4/(x- 2) is a large negative number. The limit, as x goes to 2 from below, is negative infinity. If x> 2, then, subtracting 2 from both sides, x-2> 0so x-2 is a small positive number
    4/(x-2) is a large positive number. The limit, as x goes to 2 from above, is positive infinity.

    Since "infinity" is not a number, say the limit is +/- infinity is basically a way of saying the limit "does not exist" in a particular way. saying that a limit "does not exist" does not necessarily mean it goes to +/- infinity. The situation where the limit from above is one number and the limit from below is another is a situation where the limit does not exist even though it neither one sided limit "goes to infinity".

    Either undefined or defined to be a value other than the limit. If I were to define f(x)= 1 if x is not 2, f(2)= 1000, then f has a removeable discontinuity at x= 2. The limit, as x goes to 2, is, of course, 1 but the value of the function is 1000. I could "remove" the discontinuity just by redefining f(2) to be 1.

    Well, no. f(2)= 2(8)- 4(4)+ 5(2)-4= 16-16+ 10- 4= 6

    The function must take on all values between -1 and 6 for some x between 1 and 2. That does NOT mean it cannot also take on other values. For example, if f(x)= x2 between x= -2 and x= 3, then f(-2)= 4 and f(3)= 9. In fact, the function takes on all values between 0 and 9, not just 4 and 9. Of course, 0 to 9 includes 4 to 9 so it is true that the function takes on all values between 4 and 9.

    That's because you miscalculated f(2)!

     
    Last edited: Jan 17, 2008
  6. Jan 17, 2008 #5
    ah.I didn't realise that those factors cancel.

    I think I understand this one.Division by a very small positive number->infinity.Division by a small negative number-> negative infinity

    Yes,I actually saw this in my text last night.It said as infinity is not a real number...the limit does not exist.

    So removable could be point or jump discontinuities...does this mean that point discontinuities only occur when:
    1.square roots are negative
    2.division by zero in a rational function
    3.in a piecewise function ?

    What I know about this function so far:
    f(1)=-1
    f(2)= 6
    The function is continuous on the interval [1,2]
    It's range will consist of all values between -1 and 6 for some x.

    So now that I know that it CAN take on other values besides those between(and including) -1 and 6....what is the next step??The solution on the interval [1,2]...I don't quite know what to look for...factors or an x-value or what..?

    And an unrelated question...
    I'm using a book called "Calculus,one and several variables." Salas,Hille,Etgen...
    But I am required to pass half a semester of calculus(a term).This book was definitely intended for math majors and I dont think it is aimed at the vast majority of students being introduced to calculus(unless the standard of education in South Africa is even worse than I think).Could you recommend an inexpensive,single-variable calculus book that might hold my hand a bit??
     
    Last edited: Jan 18, 2008
  7. Jan 18, 2008 #6

    HallsofIvy

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    No, I didn't say that. A "jump" discontinuity is not removable. I'm not sure what you mean by "point discontinuity". Do you mean just "removable" or simply "a discontinuity at a single point"? Most functions you will see are discontinuous only at a few points.

    Yes, but there is no "deep" mathematical fact there- just a matter of notation. Actually, "almost all" functions are badly discontinuous. But continuous functions are so easy (comparatively!) that our ways of writing functions have grown up to make it very easy to write continuous functions (and functions that are discontinuous only where they are not defined). Other functions we write by "patching together" continuous functions (piecewise).

    Well, its range includes all values between -1 and 6, as you say below. Also, you don't need that "for some x". "Some x" gives you a specific number, not the entire range.

    What was the question? You said the problem was "use the intermediate value theorem to show that there is a solution of the given equation, 2x^3 - 4x^2 +5x -4 = 0, on the interval [1,2]". That you can do just by observing the the function is continuous and that 0 is between -1 and 6. You didn't way anything about solving the equation! If you want to do that, unfortunately, this can't be factored, with integer coefficients, (the "rational root" test shows that the only possible rational number roots are [itex]\pm 1/2, \pm 1, \pm 2, \pm 4[/itex] and none of those work. There is a "cubic formula" but it is very complicated and the only other methods would be for numerical approximations.

    Actually, I rather like Salas, .... If you are only taking a semester, concentrate on the first four or five chapters. Try to solve all of the problems. You probably won't be able to do them all, but try!
     
  8. Jan 18, 2008 #7
    I have officially come right with the intermediate value theorem...I'm just wondering what I'll do when I get asked to work out something like:show that there are 3 roots on the interval [1,77]....I think I should be concerned at the moment with drilling into my subconscious the requirements for continuity and practising the intermediate value theorem.Was also working on mapping intervals,learned a lot there.Although I do have a small question to do with inequalities.

    for the inequality:(x-2)^2.(10-2x)>0

    I factored and arrived at:
    (x-2)(x-2)-5(x-5)>0

    ....assuming this is correct,what am I expected to do with respect to the -5?The inequality is supposed to be graphed on a number line.
    The examples illustrate a method;
    1.Factoring.
    2.Placing the roots on a number line.(I'm quite certain that -5>0 isn't a root,but I'd hate to have the effect of it's presence evade me!!)
    3.Substituting a value between each interval,to determine the sign of the result.


    The rational roots test and descartes rule of signs....should I be concerned with those?
    because I have this useless idiots guide to calculus which has sections on those...
     
    Last edited: Jan 18, 2008
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