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I am really struggling with Continuity

  1. Sep 9, 2015 #1
    Im uncertain if this belongs here, so I'm sorry in advance. I am extremely confused with continuity. I've tried going over my notes, reading the section, I've read about them in a separate book, I've used Khan Academy, I've talked to my professor, and I've gone to a tutor. But, to no avail on understanding them. Typically math has come very intuitively to me, but not here.

    I can understand it in terms of a graph, but solving them algebraically I am lost. The theorems confuse me with the terminology and numerous variables. I've received an A in all my previous math courses.

    Are there any resources that break these ideas down in a very elementary understanding? I'm extremely discouraged right now. Ive considered dropping. Thanks for any help.
     
    Last edited: Sep 9, 2015
  2. jcsd
  3. Sep 9, 2015 #2

    phyzguy

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    Science Advisor

    Simply put, a curve is continuous if you can draw it without lifting your pencil. What exactly is confusing you? Try to express what you don't understand, and maybe we can help.
     
  4. Sep 9, 2015 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your question is too vague; give us some examples of things you do not understand, so we have something concrete to work with. In particular, do you understand some elementary examples, but get lost when complications set in? Do you understand what the word "continuity" (or "continuous") actually mean ? Here I do not mean the so-called "epsilon-delta" stuff, but the intuitive underpinnings.
     
  5. Sep 9, 2015 #4
    It's mostly the theorems. Some seem relatively intuitive, others dont. Especially, when it comes to scatter plot graphs.

    I understand the basics of limits, as we had a brief intro to them in Pre-Calculus. The way I understand it is you are looking at when your graph gets very close to an x value, that your graph will be approaching a y value, and that y value is your limit. For example, lim x^2, x->0 = 0 because as your graph gets close to x values get closer and close to 0, your y values get closer and closer to 0.

    I know that continuity is essentially, as the aforementioned poster stated, tracing your graph without picking up your pencil. When I look at a graph, it's easy to spot whether it's continuous or not. I also know that the limit of a function is continuious if lim f(x) x->a = f(a). But, once I get the theorems thrown in, I lose concentration, and can't focus because of all the information seems overwhelming for some reason.

    But, I'll give some examples:

    1) Find lim |x|/x as x→0

    When I see see this I see that if in the denominator x ≠ 0, so I would put it as it Doesn't Not Exist (DNE). But, my professor wants us to check both the left and right hand side, and I'm confused as to why this is necessary. That is, why isn't direct substitution allowed? Why the nuances?

    Another:

    2) f(x) = (x^2+2x-3)/(x-1)

    He writes that it is discontinuous at x=1 because f(1) DNE. I see that, but, why didn't he factor the numerator? Maybe my confusion is trying to somehow mix limits and continuity together because in the previous section we were factoring to find the limit, which in this case the limit f(x) x→1 = 4, because of the hole there. But then arises the next example which is a scatter plot.

    3) f(x) = {(x^2+2x-3)/(x-1) if x ≠ 1, and 6 if x = 1

    So, here, we are asked when it is discontinuous. He factors out the numerator to find its limit is 4 as x→1, and then finds that since f(1) = 6 that it DNE. I'm confused, though. And maybe it's my notes... but if f(1) = 6 and the limit as x→1 is 4 and they are not equal, doesn't this mean there is a discontinuity at 1?

    Next, he says

    4) show f(x) = 1-√(1-x^2) is continuous on[-1, 1].

    He then proceeds to simplify the function to 1-√(1-a^2) by substitution, which he says is = to f(a). I see this. He then says this justifies (-1,1), but not [-1,1]. To justify [-1,1] he says you must check the left and right side limits. Which he checks and are both equal to 1, so he says the function has now been proven to be continuous on [-1,1].

    Why do you have to check all 3? If a function is continuous everywhere it is defined, then because the limit of 1 is continuous, and the limit of √(1-x^2) is continuous so long as the root is > 0, and since x is defined as being [-1, 1], why can't we just rationally say that the function is continuous? It seems apparent without having to check all 3?

    Lastly, this is more of a limit problem leading into continuity.

    Consider:

    5) lim x^2 sin 1/x
    x→0

    This breaks down to
    lim x→0 x^2 * lim x→0 sin 1/x,

    This is where I get confused (I'm going to shorthand and remove the x→0)σ:
    lim x^2 = 0
    lim sin 1/x = DNE

    So you have 0 * DNE

    Why don't you stop here and say the limit DNE?

    Instead he writes

    Since -1 ≤ sin 1/x ≤ 1, gotcha
    and x^2 ≥ 0, good
    then -x^2 ≤ x^2 sin 1/x ≤ x^2, HUH?

    I see what he did, but why?
    Then reasons:
    f(x)≤g(x)≤h(x) (completely lost, I know that this is from the squeeze theorem, but that's about it)

    So, lim x→0 x^2 sin 1/x = 0

    TOTALLY lost on this one.

    Thanks guys.
     
  6. Sep 9, 2015 #5

    Mark44

    Staff: Mentor

    No, this is wrong. A function f is continuous at a if ##\lim_{x \to a} f(x) = f(a)##. Here the limit is a number, so it makes no sense to say that the limit is continuous.
    Direct substitution isn't allowed because of the division by zero. The goal of this exercise is to get you to realize that the left- and right-side limits exist, and then to use those facts to say something about the two-sided limit.
    [/quote]
    Limits are very simple if you're dealing with polynomials or other functions that are continuous everywhere. The power of limits is that they can tell you something about a function at a point of discontinuity, possibly because there is division by zero at the point in question. Limits allow you to investigate the behavior of functions such as f(x) = ##\frac{x^2 - 1}{x - 1}## at x = 1 (you can't use direct substitution), as well as g(x) = ##\frac x {x^2 - 1}## near either x = 1 or x = -1.
    As you note, there is a "hole" at (1, 4), which is a removable discontinuity.
    It means that f is not continuous at x = 1. We have f being defined at 1 (i.e., f(1) = 6), and we also have ##\lim_{x \to 1} f(x) = 4##, so that limit exists. But since these two numbers are different, the function is not continous at x = 1.
    To be continuous at x = -1, you only need to check that the right-side limit equals function function value. IOW, that ##\lim_{x \to -1^+} f(x) = f(1). At the other endpoint, you only need to confirm that the left-side limit equals the function value there. The reason for using only the one-sided limits is because the domain is [-1, 1] -- the function is not defined outside this interval, so there's no point in checking limits that approach either endpoint from outside the interval.

    On its own ##\lim_{x \to 0} \sin(1/x)## does not exist. However, this expression is part of a product, and we know that ##-1 \le sin(1/x) \le 1##. Since the other factor in the product has a limit of zero, the overall product has a limit, namely zero.
    There's a theorem in your book that he's using, possibly called the "squeeze threorem" or maybe "squeeze play theorem."
     
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