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Math Help - Indeterminate Forms

  1. #16
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    Re: Indeterminate Forms

    Since the graphs of both denominators alone are continuous, wouldn't they both approach 0 from both directions?

    I guess looking from the perspective of the limit, if the limit is approaching -3 from the left, then the denominator would also be doing the same. Right?
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  2. #17
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    Re: Indeterminate Forms

    I'm considering each fraction separately.

    For the denominator, x^2+2x-3, yes, the denominator approaches 0 from the positive (above) as x approaches -3 from the negative. However, the numerator, which is x, is approaching -3, so the first fraction approaches __?__ .

    For the denominator, x+3, the graph shows that this denominator approaches 0 from below (from the negative) as x approaches -3 from the negative. The numerator is positive, so this fraction approaches __?__ .

    You subtract fractions, so your result is __?__ .
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  3. #18
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    Re: Indeterminate Forms

    Quote Originally Posted by kwikness View Post
    Right. Applying L'Hopital's rule will yield an answer of infinity. What I don't understand is which of the indeterminate forms the original equation comes out to because of the zeroes in the denominator.
    See post #12 (I corrected a typo - I referenced the wrong post). Read post #1 - the answer is clear.
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  4. #19
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    Re: Indeterminate Forms

    Quote Originally Posted by SammyS View Post
    I'm considering each fraction separately.

    For the denominator, x^2+2x-3, yes, the denominator approaches 0 from the positive (above) as x approaches -3 from the negative. However, the numerator, which is x, is approaching -3, so the first fraction approaches __?__ .

    For the denominator, x+3, the graph shows that this denominator approaches 0 from below (from the negative) as x approaches -3 from the negative. The numerator is positive, so this fraction approaches __?__ .

    You subtract fractions, so your result is __?__ .

    I think I understand now. As x approaches -3, before the denominators actually become 0, they become infinitely small, causing the fractions to evaluate to infinity.

    Therefore, I would assume the indeterminate form is:

    $\Huge\infty -\infty$

    Is this correct?
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  5. #20
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    Re: Indeterminate Forms

    What I get is (-∞)-(-∞) .
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  6. #21
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    Re: Indeterminate Forms

    But you are still better off with \frac{3x-4}{(x- 1)(x+ 3)}. Yes, the denominator still goes to 0 while the numerator goes to -13, non-zero.

    And, if x< -3, 3x- 4< -9- 4= -13, x- 1< -4, and x+ 3< 0. That is, for all x< -3, the fraction has three negative factors and so the whole fraction is negative.

    That is enough to tell you what \lim_{x\to-3^-} \left(\frac{x}{x^2+ 2x- 3}- \frac{4}{x+ 3}\right) is.
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  7. #22
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    Re: Indeterminate Forms

    The stated problem was to find what indeterminate form this was.

    I agree that to find the limit, it's best to combine the rational expressions into a single rational expression.
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  8. #23
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    Re: Indeterminate Forms

    Quote Originally Posted by SammyS View Post
    The stated problem was to find what indeterminate form this was.
    lol.. at least someone gets it. Thanks guys.
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  9. #24
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    Re: Indeterminate Forms

    Quote Originally Posted by kwikness View Post
    I think I understand now. As x approaches -3, before the denominators actually become 0, they become infinitely small, causing the fractions to evaluate to infinity.

    Therefore, I would assume the indeterminate form is:

    $\Huge\infty -\infty$

    Is this correct?
    Yes.
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  10. #25
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    Re: Indeterminate Forms

    Quote Originally Posted by kwikness View Post
    Right. Applying L'Hopital's rule will yield an answer of infinity. What I don't understand is which of the indeterminate forms the original equation comes out to because of the zeroes in the denominator.
    The original form, \lim_{x\to -3^{-}}\frac{x}{x^2+ 2x- 3}- \frac{4}{x+ 3} is of the "indeterminate form" \infty- \infty. The point of combining the fractions is to get a form that is NOT "inderminate" so you can find the limit.
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