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Value of a function definition help

  1. Mar 27, 2007 #1
    Is the value of a function defined as the limit as x approaches each value in the domain of f(x)?

    What happens if at a specific x, you have f(x)-> 0/0? is there usually a limit for which f(x) approaches in this situation?
     
    Last edited: Mar 28, 2007
  2. jcsd
  3. Mar 27, 2007 #2

    quasar987

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    Q1: no

    Q2: a limit of the form 0/infinity is equal to 0.
     
  4. Mar 28, 2007 #3
    What should the answer to Q1 be?

    WIth Q2. I am sorry, I stated it wrongly. It should be 0/0. I have corrected it in the OP.
     
  5. Mar 28, 2007 #4
    actually 0/0 is undefined. it can take different values in different situations.

    example:
    suppose f(x)= x^2/x
    when u take limit for x->0 it becomes 0/0 .
    but when u cancel x it gives ans as 0.

    suppose f(x)= x/x^2
    directly taking x-> gives again 0/0 .
    but after cancellation it gives ans as infinity.
     
  6. Mar 28, 2007 #5
    So f(0)=1 when f(x)=x/x so 0/0 in this case equals to 1.

    but 1/0 is undefined also isn't it? what is f(x)=1/x when x=0? There isn't one particular value for 1/0 like there was for 0/0 in the first situation is there? If so why?
     
  7. Mar 28, 2007 #6

    cepheid

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    Yes
    No! 0/0 is undefined. Always! It's not "equal to" any number.

    Yes.
    It's not defined for x =0. You just said so yourself.
    You're talking gibberish. 1/0 doesn't have a value. In any 'situation'.
     
  8. Mar 28, 2007 #7
    But 0/0 can in some situations like x/x=1 where you take the limit as x->0. If you define any function with the limit as x approaches each specific point in the domain than you can define situations like x/x when x=0 couldn't you?
     
  9. Mar 28, 2007 #8

    HallsofIvy

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    Your original question doesn't make sense. If x is in the domain of a function f, then, by definition of "domain", the value of the function, f(x), is defined there. Do you mean the limit of the function? It is quite possible for a function value to be defined and the limit not exist there.
    For example, f(x)= 0 if x is rational and 1 if x is irrational is defined for all real numbers but the limit does not exist anywhere.

    NO! If f(x)= x/x then its domain is "all real numbers except 0". f(0) is not defined. On the other hand, its limit as x goes to 0 IS defined and is 1. This is an example in which the limit is defined at a "limit point" of the domain that is not in the domain.

    NO again! You are confusing the limit with the value of the function. That's why your Calculus teacher is always careful to say things like
    [tex]\frac{x^2- 4}{x-2}= x+ 2[/tex] as long as x is not equal to 2.

    y= x+ 2 is defined for all x. Its graph is a straight line. (x2-4)/(x-2) is not defined at x= 2. Its graph is a straight line with a hole at (2, 4).
     
  10. Mar 28, 2007 #9
    You have raised the issue that was the reason why I created this thread. Are you claiming f(x)=x/x does not equal to g(x)=1? Hence f(x) does not equal to g(x)? If so then it means g(x) is analytic everywhere in the domain of x but f(x) is not analytic everywhere? This would also mean that if you had a function g(x)=1 then it would be wrong to multiply the top and bottom by an arbitary x.
     
    Last edited: Mar 28, 2007
  11. Mar 28, 2007 #10
    You have raised the issue that was the reason why I created this thread. Are you claiming f(x)=x/x does not equal to g(x)=1?

    f(x)=x/x = 1 = g(x) is true only when x is not equal to zero. What I (& u) calculated for other functions was the value which they approach on taking the limit. But they don't take those values actually.


    Hence f(x) does not equal to g(x)? If so then it means g(x) is analytic everywhere in the domain of x but f(x) is not analytic everywhere? This would also mean that if you had a function g(x)=1 then it would be wrong to multiply the top and bottom by an arbitary x.


    As long as x is not equal to zero u can divide it with any expression.
     
  12. Mar 28, 2007 #11
    You're being very careless. Functions are not arbitrary maps, they have DOMAINS! f(x)=x/x has natural domain (the largest one possible) of R/{0} yet g(x) has natural domain R. If two functions do not share the same domain they cannot be equal.

    f(0) is undefined until you actually define it as something. For example, we can have the function f(x)=x/x when x is not zero, and 1 when x=0 with domain R. This is the same as g(x)=1 with domain R.
     
  13. Mar 29, 2007 #12

    matt grime

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    I would suggest, until you get comfortable with this mathematically, to think of something like f(x)=x/x as a recipe, and to work out f(x) you evaluate the top, then the bottom and then divide them. You can't do that for x=0, because the function is written like that. Of course that is a contrived example.

    Better is sin(x)/x. You cannot simply by plugging in numbers evaluate that at 0. Nor can you simplify it so you can now evaluate it at 0. However, the function sin(x)/x, for x=/=0, and 1 for x=0 is a prefectly defined function on all of R (and I am abusing mathematics here horribly in an attempt to explain what's written in very stupid calculus books). Of course so is sin(x)/x for x=/=0, and 0 for x=0. One is a continuous function, the other isn't.
     
  14. Mar 29, 2007 #13

    HallsofIvy

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    g(x)=1 is obviously analytic everywhere. If by "f(x)= x/x is not analytic everywhere" in the precisely correct sense: there is at least one point at which f is not analytic- that is true. f is analytic everywhere except at 0. Yes, it would be incorrect to multiply the top and bottom of a fraction by "an arbitrary x". It would be correct to multiply the top and bottom of a fraction by any number except 0!
     
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