Zero/Zero, infinity/infinity = ?

Discussion in 'Physics & Math' started by Saint, Apr 17, 2013.

  1. Saint Valued Senior Member

    Messages:
    3,352
    Zero/Zero, infinity/infinity = ?

    Are they defined?
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. eram Sciengineer Valued Senior Member

    Messages:
    1,875
    They aren't. They're undefined.

    Please Register or Log in to view the hidden image!

     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Tach Banned Banned

    Messages:
    5,265
    You still haven't learned your prior lesson, have you? Is \(\frac{sinx}{x}\) when \(x->0\) undefined?
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Tach Banned Banned

    Messages:
    5,265
    The precise answer is given by a well-known calculus theorem.
     
  8. mathman Valued Senior Member

    Messages:
    1,427
    As is, they are undefined. If it is a result of a limiting process (like sinx/x as x -> 0), then there are procedures (L'Hopital's rule) to handle it.
     
  9. AlphaNumeric Fully ionized Moderator

    Messages:
    6,697
    Infinity is not a member of the Reals and therefore it is meaningless to try to do algebra with it unless you define the relevant rules which extend the Reals appropriately. Similarly 0/0 is meaningless because it represents 0 * (1/0), ie when you divide by X you actually multiple by the number which when multiplied by X gives 1. There is no Y such that Y*0 = 1 in the Reals and therefore 1/0 is no more a valid expression than elephant/cheese = table.
     
  10. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    Careful, Saint is easily confused.
    What eram said true: zero/zero is undefined.


    \(\lim_{x\to 0}\frac{\sin x}{x} = 1\)

    \(\frac{\sin 0}{0}\) is undefined.


    \(\lim_{x\to 0}\frac{\sin x}{x}\) is not the same as \(\frac{\sin 0}{0}\).
     
  11. AlphaNumeric Fully ionized Moderator

    Messages:
    6,697
    Exactly, which is why if you define a function \(f(x) = \frac{\sin x}{x}\) you have to define the value of f(0) separately. If you decide f(0) = 1 then that fits with the limit but that isn't required.
     

Share This Page