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Homework Help: Rules to apply L'Hospital on a limit

  1. Dec 8, 2018 at 9:21 AM #1
    1. The problem statement, all variables and given/known data
    ## \lim x-0 \frac {xcosx-log(1+x)}{x^2}##

    2. Relevant equations
    ##\frac{log(1+x)}{x}=1## ...(i)

    3. The attempt at a solution
    Using (i) we can write numerator as xcosx-x, cancelling x from denominator we have cosx-1/x, this is 0/0 form so we can use LHR which gives us -sinx/1 but this gives an incorrect value for the limit. Is there some technicality I'm overlooking about applying LHR?
     
  2. jcsd
  3. Dec 8, 2018 at 9:25 AM #2

    Math_QED

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    (i) is false. There should be a limit sign. And even then, you can't substitute a limit in another limit. Do as you are supposed to do:

    take derivative of numerator and denominator and then take the limit. If necessary, repeat.
     
  4. Dec 8, 2018 at 9:28 AM #3
    I have seen sinx/x taken to be one inside a larger limit, when can we use this and when not?
     
  5. Dec 8, 2018 at 9:29 AM #4
    That is easy enough, I'm in search of quicker methods which are practical in a time bound exam
     
  6. Dec 8, 2018 at 9:33 AM #5

    Math_QED

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    It would be applicable if you didn't make another mistake. You split up the limits like this:

    $$\lim_{x \to 0} \frac{x\cos x - \log(1+x)}{x^2} = \lim_{x \to 0} \frac{\cos x}{x} - \lim_{x \to 0} \frac{\log(1+x)}{x^2}$$

    but this is only allowed when the two limits on the right exist, and they don't. You can't apply l'Hopital's rule on the first limit on the right, as filling ##x=0## in yields ##1/0## (and it should be ##0/0## to apply that rule).

    Instead, apply l'Hopital's rule on the entire thing. Sometimes, wanting to go faster, makes you a lot slower :)
     
  7. Dec 8, 2018 at 9:37 AM #6
    True :D I suppose that's the way to go after all. Thank you very much for your help.
     
  8. Dec 8, 2018 at 1:34 PM #7

    Mark44

    Staff: Mentor

    Instead of writing \lim x-0, as you did above and in another thread, use \lim_{x \to 0}. When rendered this becomes ##\lim_{x \to 0}##.
     
  9. Dec 8, 2018 at 6:45 PM #8

    Ray Vickson

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    If l'Hospital gives ##\frac 0 0## you need to apply it again.

    Alternatively, you can expand the numerator in powers of ##x##, but you need to go to terms or at least ##x^2##; when you took ##\ln(1+x) \approx x## that was not good enough. Instead, use
    $$\ln(1+x) = x - \frac 1 2 x^2 + \cdots,$$
    where ##\cdots## stands for terms in ##x^3, x^4, \ldots## You can stop at ##\cos x = 1 + \cdots## because when you multiply by ##x## you will be looking at the main terms of order ##x##, and the missing terms will be of order higher than ##x^2.##
     
    Last edited: Dec 8, 2018 at 7:01 PM
  10. Dec 9, 2018 at 12:26 AM #9
    Oh right, thank you for informing me of the correct way to write it.
     
  11. Dec 9, 2018 at 12:28 AM #10
    Expansions can be used in almost any limit problem I see but there are about ten standard functions and all their expansions are so similar that it's easy to mix them up and end up with a negative in an exam
     
  12. Dec 9, 2018 at 2:56 AM #11
    Do you have some sort of mnemonic device though? That'd be really helpful
     
  13. Dec 9, 2018 at 4:02 AM #12

    bhobba

    Staff: Mentor

    L-Hopital is one of those things some get carried away with while others remember some simple corollaries and tricks using things like the Gamma function to do it. I tend to be a bit in between and remember a few simple corollaries - but when asked to be explicit write it out in full. The easiest corollary to remember is any polynomial divided by e^(αx) will be 0. Its like that in many areas of math - some like to remember simple 'tricks', some do it from first principles. A number of math professors mentioned it to me when doing my degree - each preferred one way or the other - its purely what feels good to you.

    As an example in the math challenge you had to evaluate (0 to ∞) ∫x^2*e^-αx . I wrote it out in full using L-Hopital, but using a couple of 'tricks' its dead simple. Do the change of variable x' = αx and you have 1/α^3 ∫x^2*e^-x = 1/α^3 Γ(3) = 2!/α^3 = 2/α^3. It's the kind of tricks you learn with practice, but I wouldn't put it on exam papers testing if you know this stuff.

    Thanks
    Bill
     
  14. Dec 9, 2018 at 4:09 PM #13

    WWGD

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    I can't think of a general one off hand, but for this last one, consider differentiating the left hand side which equals ##\frac {1}{1+x}=1-x+x^2/2-......## since the lh side is a gp with ratio x , then integrate term-by-term. A bit contrived maybe, but I can remember it.
     
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