Math Is Fun Forum

Bunno

Guest

Why does factoring work with limits????

Ok, so i want to find the limit as x approaches 1 for the equation:  X-squared - 9
                                                                                               ------------------
                                                                                                     X - 3

if i just plug in the numbers i get 0/0, but if i factorise the top first it comes out 6.

HOW IS THIS POSSIBLE?

Factorising shouldn't change the end result should it?

Bunno

Guest

Re: Why does factoring work with limits????

Sorry, i mean as X approaches 3!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Why does factoring work with limits????

"As x approaches a” means x can get as close to a as you like without being equal to a.

If your function f is continuous at a, then you can plug the value of a into f to find the limit. In the case of

is not continous at

so you cannot plug the value of 3 into

(

is not even defined at

).

However, if

then we would have

. Hence if we are interested in the value of

as

gets as close to 3 as possible without being equal to 3, then we can use

, and we see that as

approaches 3,

approaches 6.

There is a rigorous definition of limits traditionally using the symbols

and

; however I don’t think you have reached the level required to fully understand it yet, so I hope you are happy with my non-rigorous explanation above.

Last edited by JaneFairfax (2009-01-10 04:50:46)

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Why does factoring work with limits????

Until you're at the stage of understanding the rigorous definition Jane mentions above, think of getting an answer like 0/0 as 'we can't get the answer this way, as 0/0 is not defined. It could really be anything'.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Why does factoring work with limits????

No Daniel, there is a much better answer as Jane said.  To reiterate, as long as x is not 3, we are able to compute what f(x) is.  And with a limit, we are only concerned with the points around x = 3, but not x = 3 itself.  So dividing out by x - 3 is completely legit since we are making sure x is not equal to 3.

By just plugging in 3, you are actually not finding the limit at all.  Remember that f(c) = lim x->c f(x) only when f is continuous.  Since f isn't continuous at x = 3 here, plugging in 3 is very, very, very wrong.

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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: Why does factoring work with limits????

I think what Bunno was asking is "How can f(x) not be continuous at x = 3, but if you factor the numerator and cancel it with the denominator then it is continuous at x = 3".

If that was your question Bunno, then the answer is that

If you are ever solving an equation and you get (x^2 - 9) / (x - 3), then you can never allow x to equal 3, even if you factor the numerator and cancel the denominator.  You will still get x + 3 for the total expression, but you have to keep in mind that you cannot allow x to equal 3 even if the final answer you get would be continuous at that point.

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Wrap it in bacon