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Limit trouble! Help please

  1. Oct 27, 2004 #1
    I have this question from a past assignment that I never get right, it says the limit is [itex]\frac{1}{2}[/itex].
    But what I keep getting is [itex]undefined[/itex].

    Here is the problem: [itex]\lim_{x \to 1} \frac{\x^2 - 4x + 3}{\-x^2 + 2x + 3}[/itex].

    I broke it down using limit laws, then substituted in 1 for the x's, and ended up with [itex]\frac{0}{0}[/itex]. What am I doing wrong here? :confused:
     
    Last edited: Oct 27, 2004
  2. jcsd
  3. Oct 27, 2004 #2

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    Did you try doing any factoring of numerator and denominator first?
     
  4. Oct 27, 2004 #3
    Did that as well, I got [itex]\frac{(x-1)(x+4)}{(-x+1)(x+3)}[/itex]. And when you substitute 1 in, its still [itex]\frac{0}{0}[/itex].
     
  5. Oct 27, 2004 #4
    I think you factored the top wrong... I get (x-3)(x-1)... that should help you out, because I believe when you cancel (x-3) with the (x+3) on the bottom you get -1. I think.
     
  6. Oct 27, 2004 #5

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    [tex]\lim_{x \to 1} \frac{x^2 - 4x + 3}{\-x^2 + 2x + 3}[/tex]

    I am having a little bit of trouble reading your markup. Is this what we're working on?
     
  7. Oct 27, 2004 #6

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    or is that squared x term on the bottom negative?

    [tex]\lim_{x \to 1} \frac{x^2 - 4x + 3}{\ (-)x^2 + 2x + 3}[/tex]

    I am having trouble getting that negative sign to show up.
     
    Last edited: Oct 27, 2004
  8. Oct 27, 2004 #7

    This is the one. By the way, I've been meaning to ask how to make things look bigger?

    I use the itex tags, your using tex tags? Anyway, get back to me on that one later
     
  9. Oct 27, 2004 #8
    OH MY !!! :eek:
    I'm an idiot... I factored wrong... as you may have noticed, my factored numerator is magically [tex](x-1)(x+4)[/tex]... :( It should be [tex](x-1)(x-3)[/tex] right?
     
    Last edited: Oct 27, 2004
  10. Oct 27, 2004 #9
    And I think I'm wrong by including the sign...

    Should it be [tex]\frac{(x-1)(x-3)}{-(x-1)(x-2)}[/tex] rather than [tex]\frac{(x-1)(x-3)}{(-x-1)(x+2)}[/tex] ...?
     
    Last edited: Oct 27, 2004
  11. Oct 27, 2004 #10

    plover

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    For the expression you gave -(x+1)(x-3) would be the correct factoring for the denominator.

    However:
    [tex]\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 + 2x + 3)}\ =\ \frac{0}{4}\ =\ 0
    [/tex]​
    Are you sure you've got the right expression? The problem is simple unless 1 is actually a root of both the numerator and the denominator.
    [tex]\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \frac{1}{2}
    [/tex]​
    This one works out to the answer you indicated. You can either factor or use l'Hôpital's rule.
     
    Last edited: Oct 27, 2004
  12. Oct 27, 2004 #11

    plover

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    Note on tex:

    The reason the minus sign is disappearing is due to putting a backslash in front of it.
    Incorrect: [tex ]\-x^2+2x+3[/tex]

    Correct: [tex ]-x^2+2x+3[/tex]
     
  13. Oct 27, 2004 #12
    I know. I keep on double checking to see whether it wasn't just a smudge of ink or pencil mark, but yes, the sign is included...

    For your first equation, did you mean numerator instead of denominator?

    And how did you get 1/2 for the second one?

    Here's what I've been getting
    [tex]r(x) =\ \frac{(x-1)(x-3)}{-(x-1)(x-2)} =\ \frac{x-3}{-(x-2)} =\ \frac{1 - 3}{-(1-2)} =\ \frac{-2}{1}[/tex]...??

    It seems like I get farther and farther way everytime I try something else...
     
  14. Oct 27, 2004 #13

    plover

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    If your original expression is identical to the one in your book then this is a typo in the book. The limit of the original expression is 0 not 1/2.

    Also:
    Yes, I do mean the denominator. If a minus sign in front of the squared term makes the factoring difficult, put the expression in a form where you don't need to worry about it:
    -x2 - 2x + 3 = -(x2 + 2x - 3)
    The factors you give are incorrect:
    (x-1)(x-2) = x2 - 3x + 2
     
  15. Oct 27, 2004 #14
    DFDKJSDFJFLdfkdfoiaanog Wow, I always manage to read something wrong... ok but wait, doesnt [tex](x+1)(x-3) =\ x^2 - 2x - 3[/tex]?
     
  16. Oct 27, 2004 #15

    plover

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    Yes,
    (x + 1)(x - 3) = x2 - 2x - 3
    so these are not the correct factors. Keep in mind that it is necessary for 1 to be a root of the polynomial—with your factors, the roots would be -1 and 3.
     
  17. Oct 27, 2004 #16
    Alright...

    :D

    Can you show me how you did this:

    [tex]\lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \frac{1}{2}
    [/tex]

    That's with the denominator being (-x+1)(x+3) right?
     
  18. Oct 27, 2004 #17

    plover

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    Yes, that is the denominator. It's generally best though to pull the negative sign outside of the factors: -(x-1)(x+3)

    So, you've got the expression factored like this:
    [tex]
    \lim_{x \to 1} \frac{(x^2 - 4x + 3)}{(-x^2 - 2x + 3)}\ =\ \lim_{x \to 1} \frac{(x-1)(x-3)}{-(x-1)(x+3)}
    [/tex]​
    Do you see how to proceed from there?
     
    Last edited: Oct 27, 2004
  19. Oct 27, 2004 #18
    ahh... !!!!! I'm an idiot. THANKS :D
     
  20. Oct 27, 2004 #19
    I think where my confusion lies is with that damn negative sign.

    It should always be (-x)^2 -2x + 3 unless stated as -(x^2 -2x + 3)...

    Trying to make a rule for myself.. do you agree?
     
  21. Oct 27, 2004 #20

    plover

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    The standard rule is that exponents have precedence over multiplication and division, and the latter have precedence over addition and subtraction.

    So -x^2+2 without any other notations is always -(x2)+2

    This is used universally, so adopting any other convention will just lead to confusion.

    Also, to use the convention you give (i.e. addition and subtraction have precedence over exponents) consistently you would end up with:
    -x^2+2 = (-x)(2+2)
     
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