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L'Hospital proof problem

  1. Nov 9, 2012 #1
    Hey guys, first year university math student here. I need some help explaining the proof used in the scripts I'm studying from - just part of the proof to be more precise. English isn't my first language and I don't have much experience writing/rewriting down proofs and I don't know how to write those nice latex symbols, so sorry in advance if something doesn't make sense:


    Presuming:
    (1), a is element of R (|a| =/= +oo)
    (2), f and g are real functions
    (3), limit x->a_+ (f'(x) / g'(x)) exists (must be element of R, or +-oo)
    (4), limit x->a_+ (f(x)) = limit x->a_+ (g(x)) = 0

    then

    limit x->a_+ (f(x))/(g(x)) = limit x->a_+ (f'(x))/(g'(x))




    I think I understand most of the proof but there's something right at the start that I'm completely stuck at and still don't understand precisely enough:

    Let L=limit x->a_+ (f'(x) / g'(x)).

    There exists delta>0, such that for all x element of (a,a+delta), f and g are both defined on this interval,


    - I think this can be proved easily from (4), correct? Also, |f| and |g| are both smaller than some Epsilon>0. The following however, I don't understand at all:

    and both f' and g' have a finite (not = oo or -oo) derivation on this interval, and also g'=/=0.

    Why is the derivation necessarily finite?


    EDIT:

    To explain where I see the problem a bit more precisely, let's say:

    L=0
    f(x)=0 for all x element R, and therefore f'(x)=0 for all x element R

    Now, from limit x->a_+ (f'(x) / g'(x)) = 0 , it should be possible to somehow prove, that there exists a delta>0, such that for all x element (a,a+delta), g'(x) is finite and non zero. I really don't see it though, why can g'(x) not be +oo somewhere in that interval?
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2

    Stephen Tashi

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    Pick [itex] \epsilon = 0.5 [/itex] Since the above limit exists, there exists a [itex] \delta > 0 [/itex] such that [itex] a < x < a + \delta [/itex] implies [itex] | \frac{f'(x)}{g'(x) } - 0 | < 0.5 [/itex]

    The statement [itex] | \frac{f'(x)}{g'(x)}| < 0.5 [/itex] is not true unless the fraction [itex] \frac{f'(x)}{g'(x)} [/itex] exists, i.e. is a specific number with an absolute value than can be compared to 0.5. When [itex] g'(x) [/itex] is 0, the fraction doesn't exist. When [itex] g'(x) [/itex] doesn't exist by virtue of being "equal" to [itex] \infty [/itex] the fraction doesn't exist.
     
  4. Nov 10, 2012 #3
    Thanks I think I'm starting to see where the problem is -

    When [itex] g'(x) [/itex] doesn't exist by virtue of being "equal" to [itex] \infty [/itex] the fraction doesn't exist.

    Why does it not exist, if it's equal to +oo?
     
  5. Nov 10, 2012 #4

    Stephen Tashi

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    Real valued functions exist at those real numbers where their values are real numbers. [itex] \infty [/itex] is not a real number.
     
  6. Nov 10, 2012 #5
    Why does g'(x) have to be a real valued function?
     
  7. Nov 10, 2012 #6

    Stephen Tashi

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    The fraction [itex] | f'(x)/g'(x)| [/itex] isn't comparable to the real number [itex] \delta [/itex] by the relation "<" unless the fraction is a real number. The fraction isn't a real number unless it is the ratio of real numbers.
     
    Last edited: Nov 10, 2012
  8. Nov 10, 2012 #7
    Oooh. I thought that 0/oo = 0, and instead it is undefined?
     
  9. Nov 10, 2012 #8

    Stephen Tashi

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    Yes, it's undefined. Don't confuse a ratio of numbers with limit of ratios.
     
  10. Nov 11, 2012 #9
    Thank you very much!
     
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