simple question about continuity Watch

Plato's Trousers
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I am reading Mathematics : From the birth of numbers by Jan Gullberg and there is a statement I don't get. I have highlighted it on the attached image

Surely you can't say that f(x)=\dfrac{x^2-9}{x-3} is undefined at x=3 because the function is identical to f(x)=x+3

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Andythepiano
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(Original post by Plato's Trousers)
I am reading Mathematics : From the birth of numbers by Jan Gullberg and there is a statement I don't get. I have highlighted it on the attached image

Surely you can't say that f(x)=\dfrac{x^2-9}{x-3} is undefined at x=3 because the function is identical to f(x)=x+3


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If you let x=3, you will give a numerator of zero, and a denominator of zero.

Zero/zero = undefined. You always need to be on the lookout for these features
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IrrationalNumber
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Identical to x+3 except for x=3
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Plato's Trousers
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(Original post by Andythepiano)
If you let x=3, you will give a numerator of zero, and a denominator of zero.

Zero/zero = undefined. You always need to be on the lookout for these features

(Original post by IrrationalNumber)
Identical to x+3 except for x=3
But surely if you divide (x^2 - 9) by x-3 you get x+3 ? and the function x+3 has no discontinuity!

Why would you not do that first? (ie simplify the fraction before starting to stick x-values in? )

Look here

I can see why the function \dfrac{1}{x-3} for example, would be undefined at x=3, but that's because you can't cancel it down
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IrrationalNumber
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(Original post by Plato's Trousers)
But surely if you divide (x^2 - 9) by x-3 you get x+3 ?
Are you saying that 0/0=6? Rewrite your statement with x=3 and you'll see that you just said that.

and the function x+3 has no discontinuity!
Indeed but they are different at x=3.

Why would you not do that first? (ie simplify the fraction before starting to stick x-values in? )
Because you have not simplified the function, you have changed it's value at x=3

I can see why the function \dfrac{1}{x-3} for example, would be undefined
at x=3, but that's because you can't cancel it down
No, it's because you can't divide by 0
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IrrationalNumber
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It might be worth taking a look at removable singularities
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Hasufel
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it means that the function you`re given is discontinous at x=3.
-the function, expressed as it is, has a removable discontinuity at x=3, because it`s actually defined as being in the union of 2 domains: (-infin,3)union(3, +infin)

you can use L`Hopital`s rule to check what the limit should be which is limx->3 (2x/1)=6
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Plato's Trousers
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So you are saying that it's not legitimate to cancel the fraction and arrive at f(x)=x+3 ?
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smartb**d
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i don`t think he is - he`s just saying that if you were given the question: examine whether the function (x^2-9)/(x-3) is continuous - you would firstly examine it AS IS, then what you normally do in maths is see if the function can be simplified, expressed in a different form so the true nature of it can be examined. in this case, because the simplification is continuous, but the denominator of the original form isn`t,at x=3,, there`s a discontinuity as x->3 (aremovable one because we can redefine the function and domain)

(same reasoning as tan x=sinx/cosx - cos x = 0 for +-multiples of 90 degrees+2nPi (n=0,1,2,....), so tan x - the simplified function - is undefined where cosx=0 - these points are not removable discontinuities, because the simplified function is NOT DEFINED FOR THEM - original problem is STILL DEFINED for x=3
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DFranklin
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(Original post by Plato's Trousers)
So you are saying that it's not legitimate to cancel the fraction and arrive at f(x)=x+3 ?
No, it's not legitimate.

You say that (x^2-9)/(x-3) and x+3 are the same function, but they're not.

Why not? Because (x^2-9)/(x-3) isn't defined at x=3, and x+3 is, and part of the specification of a function is the domain it's defined over.

On the other hand, it makes a lot of practical sense to consider them the same thing, and the truth is "the difference is something we will treat like it's important, but actually next year when you do complex analysis, we'll call it a removable singularity and say (x^2-9)/(x-3) and x+3 are the same thing". So don't get too concerned about it - just accept it's one of the "rules of the game" for basic continuity questions.

I'll also say that I think it's quite unusual to get questions on this kind of thing that don't make it fairly explicit that (x^2-9)/(x-3) isn't defined at x = 3. (e.g. saying the function f(x) = (x^2-9)/(x-3) is defined on \mathbb{R} \setminus \{3\}).
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IrrationalNumber
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(Original post by smartb**d)
i don`t think he is
No, I am saying that they aren't the same thing and the reason for that is that x^2 - 9 over x-3 doesn't even make sense at x=3 : it's like asking for a rational square root of 2. I pretty much agree with DFranklin although I think it's worth appreciating the difference even before you study complex analysis. Indeed, I think this is worth talking about merely so you can talk about lim x goes to 3 of f(x): it does not depend on f(3) and is a well defined limit even though the value of f(3) is not defined
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Andythepiano
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DFranklin rocks - he just said everything I had been thinking with a clarity I could never accomplish
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Plato's Trousers
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(Original post by DFranklin)
No, it's not legitimate.

You say that (x^2-9)/(x-3) and x+3 are the same function, but they're not.

Why not? Because (x^2-9)/(x-3) isn't defined at x=3, and x+3 is, and part of the specification of a function is the domain it's defined over.

On the other hand, it makes a lot of practical sense to consider them the same thing, and the truth is "the difference is something we will treat like it's important, but actually next year when you do complex analysis, we'll call it a removable singularity and say (x^2-9)/(x-3) and x+3 are the same thing". So don't get too concerned about it - just accept it's one of the "rules of the game" for basic continuity questions.

I'll also say that I think it's quite unusual to get questions on this kind of thing that don't make it fairly explicit that (x^2-9)/(x-3) isn't defined at x = 3. (e.g. saying the function f(x) = (x^2-9)/(x-3) is defined on \mathbb{R} \setminus \{3\}).
Ok, many thanks for that


(Original post by Andythepiano)
DFranklin rocks - he just said everything I had been thinking with a clarity I could never accomplish
:ditto:
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ghostwalker
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Might be worth pointing out that if you specify the domain:

E.g. f(x)= x+3 \text{ on the domain }\mathbb{R}\setminus\{3\}

then it is the same function.

Though bare in mind all that's been said previous.
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