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Math Help - Limit problem

  1. #1
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    Limit problem

    Use algebra to evaluate the limit exactly.

    lim(h->0) (1/(a+h) - 1/a)/h
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  2. #2
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    Quote Originally Posted by bluejewballs View Post
    Use algebra to evaluate the limit exactly.

    lim(h->0) (1/(a+h) - 1/a)/h
    I saw you there, Jhevon. Beat you on this one . I'm gonna go back and edit now

    And here's the edit:

    Get a common denominator:


    \frac{1}{a+h} - \frac{1}{a} = \frac{a}{a(a + h)} - \frac{a+h}{(a+h)a}



    Simplify:


     = \frac{a - (a + h)}{a(a + h)} = \frac{-h}{a(a + h)}.


    Therefore:

     \frac{\frac{1}{a+h} - \frac{1}{a}}{h} = \frac{-1}{a(a + h)}.

    The limit should now be obvious.
    Last edited by mr fantastic; February 9th 2008 at 07:28 PM.
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  3. #3
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    thanks i kept getting 1/a^2 and i was told that was wrong.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bluejewballs View Post
    thanks i kept getting 1/a^2 and i was told that was wrong.
    that is wrong. it is -1/a^2

    if you have any acquaintance with calculus, you'll realize this limit gives the derivative (by its definition) of 1/a with respect to a
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  5. #5
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    Quote Originally Posted by bluejewballs View Post
    thanks i kept getting 1/a^2 and i was told that was wrong.
    So you see why the correct numerator is -1 ...?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    that is wrong. it is -1/a^2

    if you have any acquintance with calculus, you'll realize this limit gives the derivative of 1/a with respect to a
    beat me by 5 seconds .....payback time, hey?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    beat me by 5 seconds .....payback time, hey?
    exactly!
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  8. #8
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    yeah i understand that it is -1 i just made a stupid mistake like i always do
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