# Indeterminate Forms

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• Jul 28th 2011, 09:10 AM
kwikness
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?
• Jul 28th 2011, 09:42 AM
SammyS
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 __?__ .
• Jul 28th 2011, 01:43 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
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.
• Jul 29th 2011, 03:40 AM
kwikness
Re: Indeterminate Forms
Quote:

Originally Posted by SammyS
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?
• Jul 29th 2011, 06:04 AM
SammyS
Re: Indeterminate Forms
What I get is (-∞)-(-∞) .
• Jul 29th 2011, 06:17 AM
HallsofIvy
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.
• Jul 29th 2011, 06:28 AM
SammyS
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.
• Jul 29th 2011, 12:01 PM
kwikness
Re: Indeterminate Forms
Quote:

Originally Posted by SammyS
The stated problem was to find what indeterminate form this was.

lol.. at least someone gets it. Thanks guys.
• Jul 29th 2011, 02:04 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
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.
• Jul 30th 2011, 01:54 AM
HallsofIvy
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
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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