Math Is Fun Forum

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

When can this be simplified?

Hi;

I have the rational function:

This factors into:

The problem says you cannot simplify this at x= -3 because the denominator is 0, so you cannot cancel out the (x+3)

Taking the limit:

I think because the limit exists the cancellation is fine

Now at the other pole x=1/3 the limit is

So maybe at x=1/3 the cancellation is not possible. What do you think?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: When can this be simplified?

Like you said, you can cancel it everywhere except x=-3.
The function is undefined at x=-3, but it is defined everywhere close to it, so the limit of -1/10 exists.

The cancellation is allowed at x=1/3, but doing that gets you 1/0, so it's still undefined.
This time there also isn't a limit, because the limits from each direction don't agree.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: When can this be simplified?

Hi maths;

  I'm not following you here, if the limit equals -1/10 at x = -3 doesn't that mean that the function is defined at x= -3, so shouldn't I be able to do the cancellation. Sorry, I am drawing a blank here.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: When can this be simplified?

Just because the limit of f(x) as x approaches c exists does not mean that f(c) is equal to this limit.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: When can this be simplified?

Since there is a limit at x=3, you can safely cancel as long as you keep in mind that the function is undefined at that point.  More formally, you can rewrite your function as

It doesn't matter that your function is undefined at x = 1/3, since that's the case both before and after the cancellation.  You didn't change that fact by simplifying the function.

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Wrap it in bacon

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: When can this be simplified?

Thanks mathsy, Ricky and TheDude. It is a clearer now.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Riad Zaidan

Guest

Re: When can this be simplified?

Hi bobbym
From the beggining, the substitution of x=1/3 in the rational function is  1/0 so the limit is either undefined (∞) or does not exist. In our example the left hand side limit is (-∞) since the values of are <0 , and  the right hand side limit is (∞) since the values of are >0 , so the limit does not exist. Also the cancellation does not work since there is not in the numerator that can be cancelled with (3x-1) which exists in the denominator.
Best Regards
Riad Zaidan

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: When can this be simplified?

Hi Riad Zaidan;

Thanks, your right about that one too.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

rzaidan

Member

Registered: 2009-08-13

Posts: 59

Re: When can this be simplified?

Dear bobbym
thanks for your attention , and I am very pleased with you.
Best Wishes