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L'Hopital's Rule - I'm loosing my hair

  1. Apr 10, 2007 #1
    L'Hopital's Rule - I'm loosing my hair!!

    Ok, I have the following:

    Lim x->0 sqrt(4-x^2) -2 /x

    After I change the equation to remove the radicle, I get:

    Lim x->0 ((4-x^2)^1/2 - 2)/x

    but when I apply the rule the, I'm loosing it :surprised I thought I should get:

    1/2 ((4-x^2)^-1/2 times 2x)/1

    What am I doing wrong???
     
  2. jcsd
  3. Apr 10, 2007 #2

    Gib Z

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    Don't bother to change the equation, I like radicles :D

    Top, use chain rule, bottom, easy. Straight forward now.
     
  4. Apr 10, 2007 #3
    The only thing I see wrong with what you got is that the derivative of 4 - x^2 is -2x. But then again, I'm half asleep.
     
  5. Apr 10, 2007 #4
    Thanks for replying! If I use the chain rule, I get:

    1/2 sqrt(4-x^2) -2x ..... the two's cancel out but here is what I don't get (and this may be the rust showing on my math skills).... the answer shows as: -x/sqrt(4-x^2)/1 which I sort of understand. Why do I want to put the answer as: 1/2 sqrt(4-x^2) -2x with the radicle in the numerator?
     
  6. Apr 10, 2007 #5
    Well I have that

    df/dx = 1/2 ( (4-x^2)^(-1/2)(-2x) )/1
    = 1/2 (-2x) / (4-x^2)^(-1/2)
    = -x / sqrt(4-x^2)
     
    Last edited: Apr 10, 2007
  7. Apr 10, 2007 #6
    Hi Everyone,

    thanks for the replies... dimensionless - I think I know what I did wrong. As you have, I didn't raise the exponent to -1/2 thus putting the radicle in the denomonator. Now I'm getting the correct answer, but more importantly - I understand what I did wrong. THANKS for the replies!
     
  8. Apr 10, 2007 #7
    Hey - 1 last question... given that this equates to: -x / sqrt(4-x^2) I don't necessarily see that it goes to 0, unless: since -x is small and sqrt(4-x^2) is smaller, then small/smaller = 0?

    Follow me?
     
  9. Apr 10, 2007 #8
    As x -> 0, -x -> 0. As x -> 0, sqrt(4 - x^2) -> 2.

    Also, be careful about talking about equivalence here. L'Hospital's Rule just says that, if you have f(x)/g(x) such that f(x) -> 0 and g(x) -> 0 as x -> a, then f(x)/g(x) approaches the same limit as f'(x)/g'(x) as x -> a, if said limit exists.
     
  10. Apr 10, 2007 #9
    Mystic, Thanks for the clarification. That is good to note.
     
  11. Apr 10, 2007 #10
    1 more example to make sure I'm understanding this...If I have:

    lim x->1 ln x^2/(x^2+1) this can be rewitten as:

    lim x->1 (2 ln x)/(x^2+1) hence using the rule, I get:

    2 (1/x)/2x which resolves to:

    2/2x^2 and since x->1 this equals 1

    Am I getting this right?
     
  12. Apr 10, 2007 #11

    VietDao29

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    Nope, that's not of any Indeterminate Forms. You should note that L'Hopital's rule can only be applied to the 2 Indeterminate Forms [tex]\frac{0}{0}[/tex] and [tex]\frac{\infty}{\infty}[/tex]. In this problem, we have:
    [tex]\lim_{x \rightarrow 1} \ln (x ^ 2) = \ln (1 ^ 2) = 0[/tex]
    and:
    [tex]\lim_{x \rightarrow 1} (x ^ 2 + 1) = 1 ^ 2 + 1 = 2[/tex], so your limit should be:
    [tex]\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 + 1} = \frac{0}{2} = 0[/tex]
    You are over-complicating the problem. :rolleyes:
     
  13. Apr 10, 2007 #12
    VietDao, I had a typo: it should have read:

    lim x->1 ln x^2/(x^2<b>-</b>1) Sorry for the error. I followed what you had and that makes sense so I think I'm getting it albeit slowly. Using the revised problem, I would get 0/0?
     
  14. Apr 10, 2007 #13

    VietDao29

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    Yes, if it does read:
    [tex]\lim_{x \rightarrow 1} \frac{\ln (x ^ 2)}{x ^ 2 - 1}[/tex], then your solution is totally correct. :)
     
  15. Apr 10, 2007 #14
    :-)

    Thanks for the assistance everyone!!!!
     
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