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L'Hospital proof problem 
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#1
Nov912, 08:01 AM

P: 11

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? 


#2
Nov912, 07:19 PM

Sci Advisor
P: 3,313

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. 


#3
Nov1012, 03:22 AM

P: 11

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? 


#4
Nov1012, 09:07 AM

Sci Advisor
P: 3,313

L'Hospital proof problem
Real valued functions exist at those real numbers where their values are real numbers. [itex] \infty [/itex] is not a real number.



#5
Nov1012, 10:42 AM

P: 11




#6
Nov1012, 02:00 PM

Sci Advisor
P: 3,313

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.



#7
Nov1012, 08:21 PM

P: 11

Oooh. I thought that 0/oo = 0, and instead it is undefined?



#8
Nov1012, 08:54 PM

Sci Advisor
P: 3,313




#9
Nov1112, 04:52 AM

P: 11

Thank you very much!



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