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Basic limits of rational functions: behavior near vertical asymptotes

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    We are required to sketch a (reasonably accurate) picture of a rational function f(x) = P(x)/Q(x) with P, Q polynomials in x and Q nonzero. We know that the roots of Q(x) are, say, x1, x2, etc. and so f(x) is (typically) asymptotic to the vertical lines x = xk for each k.

    We want [tex]\lim_{x \rightarrow {x_{k}}^{\pm}} f(x)[/tex] for each xk. Note that we need the limiting behavior of f(x) for x < xk and x > xk, separately, in order to accurately sketch the behavior of f(x) near these vertical asymptotes. That's a total of 2k limits we need to solve, though they are each very similar.

    I'm a tutor and so I need to present the solution of this to my student. But it kinda sucks that I don't know how to evaluate this kind of simple one-sided limit problem off hand. I'm pretty sure that there's some algebra trick we can use to express the rational function in a different form that makes the limit easy to evaluate using the typical "limit laws".

    I don't want to have to go all ϵ-δ on my student though :p He knows the formal definition of "limit", but I feel that we don't need to refer to it to evaluate limits of rational functions because we've already proven the limit laws.

    For example, say [tex]f(x) = {x^3 - 3x^2 - 4x + 12 \over x^2 - 1}[/tex] Since the roots of x2 - 1 are x = ±1, we want the following limits:

    [tex]\lim_{x \rightarrow {1}^{+}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1} \qquad \lim_{x \rightarrow {1}^{-}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1}[/tex]
    [tex]\lim_{x \rightarrow {(-1)}^{+}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1} \qquad \lim_{x \rightarrow {(-1)}^{-}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1}[/tex]

    3. The attempt at a solution

    [tex]\lim_{x \rightarrow {1}^{+}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1} = \infty[/tex] To show this, I suppose I can argue by evaluating the expression for numbers close to (but greater than) x = 1 and observing the quantity is large and positive. So we can "plug in" x = 1.1, 1.01, 1.001, etc. into the limiting expression and see that it grows without bound. But this is not very rigorous/formal at all.

    Also, this is just a "sub-problem" in the larger problem of trying to sketch the graph of this curve. I don't want to have to make these kinds of tedious numerical arguments when presenting the reasoning here; it will just get in the way of the larger problem.

    Isn't there a better way to quickly have the solution here? I feel like this a pretty typical elementary limit problem.
     
    Last edited: Aug 9, 2011
  2. jcsd
  3. Aug 9, 2011 #2
    Expand it using partial fractions. To do that, first use long division.
     
    Last edited: Aug 9, 2011
  4. Aug 9, 2011 #3
    ^ How do you do that when the degree of the P(x) is greater than that of Q(x)?
     
  5. Aug 9, 2011 #4
    ....
     
  6. Aug 9, 2011 #5
    Okay I just suck at reading. I'm still stuck though, honestly your post is too vague to help. Can you be more specific to the example I gave above?
     
  7. Aug 9, 2011 #6
    [PLAIN]http://img26.imageshack.us/img26/9818/longdivision07.gif [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Aug 9, 2011 #7
  9. Aug 9, 2011 #8
    @Harrisonized: Great. Why are you even posting if you're just going to be a dick?

    @Bohrok: Well, I already know how to do that. I just don't see how it will help.

    Anyway, it's not like I didn't try division or partial fractions before you mentioned that. I get this far:

    [tex]{x^3 - 3x^2 - 4x + 12 \over x^2 - 1} = x - 3 - {3 \over {x + 1}} + {6 \over {(x + 1)(x - 1)}}[/tex]

    Writing the last term in partial fractions and simplifying:

    [tex]{x^3 - 3x^2 - 4x + 12 \over x^2 - 1} = x - 3 + {3 \over {x - 1}} - {6 \over x + 1}[/tex]

    Now what do I write, specifically, to show

    [tex]\lim_{x \rightarrow 1^{+}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1} = \lim_{x \rightarrow 1^{+}} (x - 3 + {3 \over {x - 1}} - {6 \over x + 1}) = \infty[/tex]

    Do I have to use a trick like, let [itex]t = x - 1[/itex] so that [itex]t \rightarrow 0^+[/itex] as [itex]x \rightarrow 1^+[/itex]? Even then, won't I end up with an expression such as [itex]1 - \infty[/itex]? It seems really informal to just claim that this is the same as [itex]\infty[/itex].

    Even in a simpler example such as [itex]\lim_{x \rightarrow 0^+} ({x^2 - 1 \over x})[/itex], we can write [itex]{x^2 - 1 \over x} = {x - {1 \over x}}[/itex] and still be ****ed to write something informal like "when x is small and x > 0, we have a small number minus a large number, which is a large negative number, so [itex]\lim_{x \rightarrow 0^+} {x - {1 \over x}} = -\infty[/itex]".
     
    Last edited: Aug 9, 2011
  10. Aug 9, 2011 #9
    Well usually with problems like this, I would make intervals using zeros and points where the function is undefined as the boundaries between them, then use test points to determine whether that interval is positive or negative; I haven't tried it on this problem, but it should help.

    One thing about doing the long division is that you can see that the last two terms tend to 0 as x → ∞,
    [tex]x - 3 + {3 \over {x - 1}} - {6 \over x + 1} \approx x - 3[/tex]
    for large x, so you know the function kind of looks like x - 3.
     
  11. Aug 9, 2011 #10
    Because you're a tutor. You're supposed to know this stuff unless you want to screw over your students. And unless my eyes deceive me, you also say you're an undergrad in pure mathematics. This is the kind of material people learn long before delta-epsilon limit definitions, and it has parallels to elementary school days.

    Now listen up:

    Anything + ∞ = +∞
    Anything - ∞ = -∞

    ∞ is so large that it makes everything else negligible. Only ∞ can match itself, and when that happens, you either get ∞ again or you get an indeterminate form.
     
  12. Aug 9, 2011 #11
    "Anything + infinity" = "infinity"? That's exactly the kind of informality I'm trying to avoid here. I know what you just said about "anything plus infinity" is intuitive. The question I've been asking since the OP was how to make that a little less silly.

    Of course I learned this far before I learned epsilon-delta definitions. But I was presented this in its highschool-esque "anything + infinity = infinity" form, which is not the way I want to present it to my student. You don't have to tell me that I should be prepared for my students. I know that. That's why I'm asking about here, it so I can be prepared when I teach.
     
  13. Aug 9, 2011 #12

    Hurkyl

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    It's not silly. Not only is this the way you should be thinking about it, but it's literally true. (also important is that x+y is continuous at [itex](x,y) = (a,+\infty)[/itex] whenever [itex]a \neq -\infty[/itex])

    Introducing the extended real numbers is one of the first things a real analysis text would do. (and honestly, I'm somewhat perplexed as to why they're not introduced in one's first calculus class if not before)

    Even if you don't use the extended real numbers, you would still mimic them with limit laws, such as
    If [itex]\lim f(x) = a[/itex] (for a finite a) and [itex]\lim g(x) = +\infty[/itex], then [itex]\lim f(x) + g(x) = +\infty[/itex]​


    Honestly, how to use estimates and approximations are one of the major skills one is supposed to learn in calculus. You already split the fraction into parts and completely analyzed each term. (three are bounded, and one converges to [itex]+\infty[/itex]) -- the rest of the argument is very short. Something like:

    Let N > 0.
    Choose delta so that if 1 < x < 1 + delta, then the big term is bigger than N + 100, and the other terms are each individually less than 10 in magnitude. Thus, their sum is bigger than N.
     
    Last edited: Aug 9, 2011
  14. Aug 9, 2011 #13
    This makes more sense. I could stop to prove this result and then move on. Still, without the hyperreals, statements like "one minus infinity" don't really make much sense, formally speaking. That's what I meant by "silly". We're talking about an elementary calculus student, and a science major at that, so there's no point introducing something like the extended reals. I've already pushed it by bothering to show him epsilon-delta definitions instead of making the limit an intuitive statement. But by that logic, an argument like "a fixed positive number minus a large positive number is a large negative number" should suffice then.

    I taught epsilon-delta to this particular student, but I meant that for this one portion of the problem, I didn't want to have to break it up into small bits that each take a (relatively) long time to show, i.e. I'd have to find the limit from first definitions. A tool that makes it very simple to take the one-sided limits in this problem would be very useful.
     
  15. Aug 10, 2011 #14

    Hurkyl

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    The hyperreals are not what I was talking about....

    The extended reals are part of elementary calculus. The only question is whether you acknowledge them as a number system, or if you find ad hoc ways to talk around them -- e.g. talking about "limit forms", or replicating their arithmetic in the form of limit laws.




    You're doing it wrong. You don't find the small bits from first definitions -- you find them in efficient ways. Then you use first definitions to convert that calculation into something you can use.

    First, compute [itex]\lim_{x \to 1} x - 3 - 6/(x+1) = -5[/itex] (by your favorite means)
    Now, because we have this fact, for any value A < -5 we can say there exists a [itex]\delta[/itex] such that
    If [itex]1 < x < 1 + \delta[/itex], then [itex]A < x - 3 - 6/(x+1)[/itex]​

    Second, compute [itex]\lim_{x \to 1^+} 3/(x-1) = +\infty[/itex] (by your favorite means)

    Now, let N be a real number.
    By the above computations, there exists [itex]\delta > 0[/itex] such that
    • If [itex]1 < x < 1 + \delta[/itex], then [itex]-10 < x - 3 - 6/(x+1)[/itex]
    • If [itex]1 < x < 1 + \delta[/itex], then [itex]N + 10 < 3/(x-1)[/itex]
    Adding the inequalities further implies
    If [itex]1 < x < 1 + \delta[/itex], then [itex]N < {x^3 - 3x^2 - 4x + 12 \over x^2 - 1}[/itex]
    And this is exactly what we needed to do to prove [itex]\lim_{x \rightarrow {1}^{+}} {x^3 - 3x^2 - 4x + 12 \over x^2 - 1} = +\infty[/itex].


    (I skipped a step -- there's a delta that makes the first bullet point true, and there's another delta that makes the second bullet point true. But if you take the minimum of the two, that's a delta that makes both bullet points true)​
     
    Last edited: Aug 10, 2011
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