1. ## 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?

2. ## Re: Indeterminate Forms

I'm considering each fraction separately.

For the denominator, $\displaystyle x^2+2x-3$, 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, $\displaystyle x+3$, 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 __?__ .

3. ## Re: Indeterminate Forms

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.

4. ## Re: Indeterminate Forms

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

For the denominator, $\displaystyle x^2+2x-3$, 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, $\displaystyle x+3$, 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:

$\displaystyle$\Huge\infty -\infty$$Is this correct? 5. ## Re: Indeterminate Forms What I get is (-∞)-(-∞) . 6. ## Re: Indeterminate Forms But you are still better off with \displaystyle \frac{3x-4}{(x- 1)(x+ 3)}. Yes, the denominator still goes to 0 while the numerator goes to -13, non-zero. And, if x< -3, 3x- 4< -9- 4= -13, x- 1< -4, and x+ 3< 0. That is, for all x< -3, the fraction has three negative factors and so the whole fraction is negative. That is enough to tell you what \displaystyle \lim_{x\to-3^-} \left(\frac{x}{x^2+ 2x- 3}- \frac{4}{x+ 3}\right) is. 7. ## Re: Indeterminate Forms The stated problem was to find what indeterminate form this was. I agree that to find the limit, it's best to combine the rational expressions into a single rational expression. 8. ## Re: Indeterminate Forms Originally Posted by SammyS The stated problem was to find what indeterminate form this was. lol.. at least someone gets it. Thanks guys. 9. ## Re: Indeterminate Forms Originally Posted by kwikness I think I understand now. As x approaches -3, before the denominators actually become 0, they become infinitely small, causing the fractions to evaluate to infinity. Therefore, I would assume the indeterminate form is: \displaystyle \Huge\infty -\infty$$

Is this correct?
Yes.

10. ## Re: Indeterminate Forms

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, $\displaystyle \lim_{x\to -3^{-}}\frac{x}{x^2+ 2x- 3}- \frac{4}{x+ 3}$ is of the "indeterminate form" $\displaystyle \infty- \infty$. 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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