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Ok, so i want to find the limit as x approaches 1 for the equation: X-squared - 9
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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?
Sorry, i mean as X approaches 3!
"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 dont 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)
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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'.
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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.
"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..."
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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.
Wrap it in bacon
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