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Troube with limits

  1. Mar 4, 2006 #1
    Sorry... But I am always struggling with limits...
    This time, it's this
    [tex]{\lim_{x \rightarrow 1} \frac{1}{(x-1)^2} - \frac{ax}{(x^2+1)^2}[/tex]
    a is a parameter, so we have to discuss the limit when a equals different values. Am I right? if so, how can we do that?

    This one is similar too, but I think if I know how to do the first one, I'll be able to do this one too:
    [tex]\lim_{x \rightarrow -1} \frac{(x-a)(x-b)}{x^2+x}[/tex]
    but do we have to discuss both a and b ??
    Thank you a lot!
    Last edited: Mar 4, 2006
  2. jcsd
  3. Mar 4, 2006 #2
    In order to discuss the limit for different values of a and b you would just give an answer in terms of a or b, an answer like 4*a or 20b/a.

    However, in these examples neither the a or the b would b significant. Look at the denominator of both equations and understand how they work as x approaches the intended number?

  4. Mar 5, 2006 #3


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    The first one, at least, is straight forward: what is
    [tex]{\lim_{x \rightarrow1} \frac{ax}{(x^2+1)^2}[/tex]

    What happens to [tex]\frac{1}{(x-1)^2}[/tex]
    as x goes to 1?

    The second one is quite different. It will help to factor x2+ x as x(x+ 1). What happens when x is -1? In order for that to have a limit, what has to happen in the numerator?

    I don't think your answer should be a function of a and b like lyuokdea said. It will be more like "the limit does not exist unless a or b equal ___ and then it is___".
    Last edited by a moderator: Mar 5, 2006
  5. Mar 5, 2006 #4
    lim ax/(x²+1)² is a/2x
    1/(x-1)² approaches to infinity as x goes to 1

    second one, the numerator shouldn't be equal to zero, so x should be different than -1
  6. Mar 5, 2006 #5


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    Yes, 1 / (x - 1)2 will tend to infinity as x tends to 1.
    And no, [tex]\lim_{x \rightarrow 1} \frac{ax}{(x ^ 2 + 1) ^ 2} = \frac{a}{2 ^ 2} = \frac{a}{4}[/tex], not a / 2x, as you said.
    Now if one part tends to infinity (i.e, some very very great number), and the other tends to a / 4 (which is obviously a number), then what can you say about this limit, does it converge to any value? Or is it unbounded?
    No, think about it, if the numerator tends to some number that's not 0, then does the limit exist? For example: Does this limit exist:
    [tex]\lim_{x \rightarrow 1} \frac{x}{x - 1}[/tex]? In the example, the numerator tends to 1, while the denominator tends to 0.
    So in order for this expression: [tex]\lim_{x \rightarrow -1} \frac{(x - a) (x - b)}{x ^ 2 + x}[/tex] to have limit, what should the numerator go to?
    Can you follow me? :)
  7. Mar 5, 2006 #6
    Okay, so the first one is infinity :)
    For the second one, it's undefined. In order for it to be defined, a and b should'nt equal -1 ?
  8. Mar 5, 2006 #7
    The point of the second one is that the only possible values of a and b for the limit to exist are those that make your limit 0/0. That's not to say that the limit DOES exist, just that these are the only ones where the possibility exists. So what values of a and b make the numerator zero in the limit? And what, in fact, is the limit if a and b take those values?

  9. Mar 6, 2006 #8


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    This one is correct. :)
    However, this one is just so wrong... Why shouldn't they equal -1?
    In fact, as I said earlier, a fraction, whose numerator tends to some value that is not 0, and denominator tends to 0, then the limit does not exist.
    For example:
    [tex]\lim_{x \rightarrow 1} \frac{x}{x - 1}[/tex]. As x goes to 2, the numerator will tend to 1, and the denominator will tend to 0. Hence, there's no limit there. Why? It's because the expression will become bigger, and bigger as x tends towards 1.
    So in order for the limit:
    [tex]\lim_{x \rightarrow -1} \frac{(x - a) (x - b)}{x ^ 2 + x}[/tex] to exist, then the numerator must also tend to 0, right? And it'll become one of the Indeterminate forms, namely 0 / 0, and you can just solve it from there.
    Now as x tends towards -1, you must have (x - a)(x - b) -> 0, so what's a, and b?
    Can you go from here? :)
  10. Mar 6, 2006 #9
    so either a should equal -1, or b should equal -1, or both of them should equal -1. Then we'll get an undetermined form of 0/0. But what can we do to avoid this undetermined form ?
  11. Mar 6, 2006 #10


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    Yes, correct. :)
    However, we don't "avoid" the indeterminate forms, we should try to solve it. :)
    I'll start off for you.
    If a = -1, the whole expression becomes:
    [tex]\lim_{x \rightarrow -1} \frac{(x + 1) (x - b)}{x ^ 2 + x} = \lim_{x \rightarrow -1} \frac{(x + 1) (x - b)}{x (x + 1)} = \lim_{x \rightarrow -1} \frac{x - b}{x}[/tex].
    It's common to factor both numerator, and denominator, then try to cancel out as much terms that tend to 0 as possible. In this example, as x tends to -1, (x + 1) tends to 0. And the 2 terms that tend to 0 (i.e (x + 1)) have cancelled themselves out, and the limit is no longer in any Indeterminate form.
    Can you go from here? Just do the same for b. :)
    By the way, it's Indeterminate form, not undetermined form. :)
  12. Mar 6, 2006 #11
    Yup!! Thank you a lot :)
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