# Indeterminate Forms

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jul 27th 2011, 03:48 PM
kwikness
Indeterminate Forms
Hi guys, I'm working through a bunch of problems in which I have to determine the type of indeterminate form (i.e. infin/infin, 0/0). This one is throwing me for a loop though:

$\underset{x\to -{3}^{-}}{lim}\left(\frac{x}{{x}^{2}+2x-3}-\frac{4}{x+3}\right)$

When I substitute in the -3, I get 3/0 - 4/0, but I obviously can't divide by 0. What is the indeterminate form in this case?
• Jul 27th 2011, 04:07 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
Hi guys, I'm working through a bunch of problems in which I have to determine the type of indeterminate form (i.e. infin/infin, 0/0). This one is throwing me for a loop though:

$\underset{x\to -{3}^{-}}{lim}\left(\frac{x}{{x}^{2}+2x-3}-\frac{4}{x+3}\right)$

When I substitute in the -3, I get 3/0 - 4/0, but I obviously can't divide by 0. What is the indeterminate form in this case?

Start by expressing it all over a common denominator and then simplify. Note that x^2 + 2x - 3 = (x+3)(x-1) ....
• Jul 27th 2011, 04:18 PM
kwikness
Re: Indeterminate Forms

I've tried simplifying it down to:

$\frac{3x-4}{(x-1)(x+3)}$

But still, when I plug in -3, I get -13/0.
• Jul 27th 2011, 04:19 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness

I've tried simplifying it down to:
$\frac{3x-4}{\left(I-x\right)\left(x+3\right)}$

But still, when I plug in -3, I get -13/0.

Yes, so what does that tell you ....? What would a graph of the function look like? (NOte that the form is no longer indeterminant ....)
• Jul 27th 2011, 04:20 PM
pickslides
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
Hi guys, I'm working through a bunch of problems in which I have to determine the type of indeterminate form (i.e. infin/infin, 0/0). This one is throwing me for a loop though:

$\underset{x\to -{3}^{-}}{lim}\left(\frac{x}{{x}^{2}+2x-3}-\frac{4}{x+3}\right)$

When I substitute in the -3, I get 3/0 - 4/0, but I obviously can't divide by 0. What is the indeterminate form in this case?

Maybe this will help,

$\displaystyle \lim_{x\to -3}\left(\frac{x}{{x}^{2}+2x-3}\times \frac{\frac{1}{x^2}}{\frac{1}{x^2}}-\frac{4}{x+3}\times \frac{\frac{1}{x}}{\frac{1}{x}}\right)$
• Jul 27th 2011, 04:25 PM
kwikness
Re: Indeterminate Forms
It looks like there's a vertical asymptote in the graph, but the graph runs through the x axis at (-3,0)

Does that mean the correct answer is that this is NOT an indeterminate form?
• Jul 27th 2011, 04:31 PM
kwikness
Re: Indeterminate Forms
pickslides, your avatar is the balls, but I can't make sense of how your hint helps as those additions are really just 1s and I'm left with the same equation when I plug in the -3.
• Jul 27th 2011, 06:42 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
It looks like there's a vertical asymptote in the graph, but the graph runs through the x axis at (-3,0)

Does that mean the correct answer is that this is NOT an indeterminate form?

-13/0 => the limit will be either +oo or -oo. You need to figure out which.
• Jul 27th 2011, 07:53 PM
SammyS
Re: Indeterminate Forms
Quote:

Originally Posted by kwikness
Hi guys, I'm working through a bunch of problems in which I have to determine the type of indeterminate form (i.e. infin/infin, 0/0). This one is throwing me for a loop though:

$\underset{x\to -{3}^{-}}{lim}\left(\frac{x}{{x}^{2}+2x-3}-\frac{4}{x+3}\right)$

When I substitute in the -3, I get 3/0 - 4/0, but I obviously can't divide by 0. What is the indeterminate form in this case?

What are all of the indeterminate forms you have learned?

How does $\lim_{x\to-3^-}$ differ from $\lim_{x\to-3^+}\,?$

It may help to factor $x^2+2x-3\,,$ but I think it's a mistake to combine the fractions into one fraction.
• Jul 27th 2011, 09:54 PM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by SammyS
[snip]

It may help to factor $x^2+2x-3\,,$ but I think it's a mistake to combine the fractions into one fraction.

I don't see why it would be a mistake. Doing so makes it very clear that it will be one of -oo or +oo.
• Jul 28th 2011, 02:39 AM
SammyS
Re: Indeterminate Forms
Quote:

Originally Posted by mr fantastic
I don't see why it would be a mistake. Doing so makes it very clear that it will be one of -oo or +oo.

Yes, this will give the desired limit, but the assignment was to determine the indeterminate form of the original expression.

Some possible indeterminate forms: 0/0, ∞/∞, 0·∞, ∞ - ∞, -∞ + ∞, 1^∞, ∞^0, 0^0 etc
• Jul 28th 2011, 05:08 AM
mr fantastic
Re: Indeterminate Forms
Quote:

Originally Posted by SammyS
Yes, this will give the desired limit, but the assignment was to determine the indeterminate form of the original expression.

Some possible indeterminate forms: 0/0, ∞/∞, 0·∞, ∞ - ∞, -∞ + ∞, 1^∞, ∞^0, 0^0 etc

Well, it is one of the forms in your list. And post #1 (of all posts, the original question) suggests what the answer is ....
• Jul 28th 2011, 05:41 AM
SammyS
Re: Indeterminate Forms
Yes.

However, it appears that kwikness (OP) does get what happens as the denominator approaches zero.
• Jul 28th 2011, 08:09 AM
kwikness
Re: Indeterminate Forms
Quote:

Originally Posted by SammyS
Yes.

However, it appears that kwikness (OP) does get what happens as the denominator approaches zero.

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.
• Jul 28th 2011, 08:36 AM
SammyS
Re: Indeterminate Forms
Do the denominators approach 0 from the right (the positive side of zero) or from the left (the negative side) as x approaches -3 from the left?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last