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?

I'm considering each fraction separately.

For the denominator, , 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, , 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 __?__ .

Quote:

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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.

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See post #12 (I corrected a typo - I referenced the wrong post). Read post #1 - the answer is clear.

Quote:

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Originally Posted by SammyS

I'm considering each fraction separately.

For the denominator, , 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, , 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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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:



Is this correct?

What I get is (-∞)-(-∞) .

But you are still better off with . 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 is.

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.

Quote:

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Originally Posted by SammyS

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

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lol.. at least someone gets it. Thanks guys.

Quote:

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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:



Is this correct?

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Yes.

Quote:

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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.

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The original form, is of the "indeterminate form" . The point of combining the fractions is to get a form that is NOT "inderminate" so you can find the limit.