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L ' Hopital's Rule for x -> infinity

  1. Jul 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Hopital's Rule for x -> ∞ applies the same way as x -> 0.


    2. Relevant equations

    As shown in attached pic.

    3. The attempt at a solution

    I tried to prove that x->∞ works the same way as x -> 0, only to get its reciprocal.. Not sure what is wrong with my working..
     

    Attached Files:

  2. jcsd
  3. Jul 6, 2012 #2
    Why should h or k have a derivative at a? You'll have to come up with different reasoning.
     
  4. Jul 6, 2012 #3

    HallsofIvy

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    You ask about L'Hopital's rule applying "as x-> infinity" but your attachment has x->a while it is f(x) and g(x) that go to infinity. Which are you trying to prove?
     
  5. Jul 6, 2012 #4

    Ray Vickson

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    No only that, you applied Taylor expansions of f(x) and g(x) at x = a, where both functions are infinite! You can't do that.

    RGV
     
  6. Jul 6, 2012 #5
    Sorry for the confusion! I mean f(a) = g(a) = ∞

    I earlier showed that l'Hopital's Rule works when its 0/0 but I am now trying to show it works for ∞/∞.

    So i took (1/g)/(1/f) to make it 0/0..

    But when i use taylor's expansion something's wrong..
     
  7. Jul 6, 2012 #6

    SammyS

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    What is Taylor's expansion for 1/(f(x)) , expanded about x = a ?
    This poses a problem since 1/f(a) is undefined.​

    Construct two functions, F(x) & G(x) such that F(a) = 0 = G(a), otherwise F(x) = 1/f(x) and G(x) = 1/g(x) .
     
  8. Jul 9, 2012 #7
    Yes i did that, i let 1/g = h, 1/f = k.

    then it reduces to [ h'(x)/k'(x) ] which translates to [g'(x)/f'(x)], which is the reciprocal instead..
     
  9. Jul 9, 2012 #8

    SammyS

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    Check that again.

    So, [itex]\displaystyle h(x)=\frac{1}{g(x)}\quad\to\quad h'(x)=-\frac{g'(x)}{(g(x))^2}\ .[/itex]

    A similar result holds for k'(x) .

    [itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{h(x)}{k(x)}[/itex]
    [itex]\displaystyle =\lim_{x\to a}\frac{h'(x)}{k'(x)}[/itex]

    [itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{(g(x))^2}\frac{(f(x))^2}{f'(x)}[/itex]

    [itex]\displaystyle =\lim_{x\to a}\frac{g'(x)}{f'(x)}\ \lim_{x\to a}\frac{(f(x))^2}{(g(x))^2}[/itex]​

    Can you take it from there?
     
  10. Jul 10, 2012 #9
    Yes I ended up with that step. So f(a) = g(a), f/g = 1.

    Then we have g'(x)/f'(x), which is the reciprocal, where I am stuck at...
     
  11. Jul 10, 2012 #10

    SammyS

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    What is that reciprocal equal to ?
     
  12. Jul 10, 2012 #11
    By right I should get f'(x)/g'(x) as when approximating x→0.

    But i'm getting g'(x)/f'(x) instead..
     
  13. Jul 10, 2012 #12

    SammyS

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    Suppose both limits [itex]\displaystyle \lim_{x\to a}f'(x) \,,\ \lim_{x\to a}g'(x)[/itex] exist & that [itex]\displaystyle \lim_{x\to a}f'(x)\ne0\ .[/itex]

    Then the following should give the desired result.
    [itex]\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}=
    \frac{\lim_{x\to a}g'(x)}{\lim_{x\to a}f'(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}\cdot \lim_{x\to a}\frac{f(x)}{g(x)}
    [/itex]​
     
  14. Jul 11, 2012 #13
    I see! then the f(x)/g(x) cancels on both sides leaving only one f(x)/g(x) on the right, then you bring over the g'(x)/f'(x) to the left to give f'(x)/g'(x) as desired!
     
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