From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: a non-mathematician asks... Date: 11 Feb 1999 13:42:52 -0500 Newsgroups: sci.math In article <01be5557$4ed4fed0$047ec997@drdave>, David McCutchan wrote: : :L'hospital figured out that expressions like yours : :Limit (x approaches 0) sinx /x = 0/0 are solved by figuring out which :function is approaching zero faster. "Fast" means a rapid change in :derivative, : :So his rule says (1) differentiate the numerator and denominator :individually and (2) again examine the limit as x =0 : :So we have cos x/1 = cos 0/1 = 1. : (1) Mathematical reservation: This works as stated for functions known to be expandable in power series around the point (0 in this case) in the first place. It is then also of interest in complex analysis. For functions of real variable, there is a logical turn-around, which often looks too subtle to the untrained: We check that f(x) and g(x) both go to zero, or that g(x) goes to infinity, then we try to find the limit of f'(x)/g'(x), and if the limit is found to be L, then the limit of f(x)/g(x) is the same number L. (Examples exist when lim f(x)/g(x) exists, f and g go to 0, yet f'(x)/g'(x) fails to have a limit. ) The proof of the "infinity/infinity" case is quite tough, definitely not for Year I Calculus. (2) Historical trivium: The Marquis de l'Hospital (1661-1704) published the rule in his Calculus textbook; he was an accomplished textbook author for his times. But the authorship belongs to a Swiss mathematician Johann Bernoulli (1667-1748), who sold the discovery to l'Hospital. Cheers, ZVK(Slavek). ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: L'hopital's rule Date: 26 Oct 1999 01:06:54 -0400 Newsgroups: sci.math Keywords: Proof of L'Hospital's Rule In article , Luc Taylor wrote: :My calculus book is irritating me! It says that a more general version :of L'hospital's rule can be found with accompanying proofs in more :advanced books. I don't think they should present anything that they :can't supply with a proof. : :Please if anyone knows of a complete copy of L'hospital's rule with an :accompanying proof, please email it to me. : One version is in Rudin's book Principles of Mathematical Analysis, Theorem 5.13. I have another, attached. By the way, the 0/0 case is handled by Mean Value Theorem; I presume your textbook presented that one. This is the "hard L'Hospital's Rule, infinity/infinity. Cheers, ZVK(Slavek). The Hard L'Hospital Rule. (Paraphrased from a Slovak textbook.) A special case. --------------- Let let G(x) -> inf, and let lim F'(x)/G'(x) = 0 as x -> inf. Then lim F(x)/G(x) = 0. Proof: To every eps > 0, we want to find x0 such that if x >= x0 then -eps < F(x)/G(x) < eps. Let eps > 0. In what follows, we will repeat raising the lower bound for x (from x1 to x2 etc.) to impose a new condition, and each time all previous conditions will be satisfied. ---> Since G(x) -> inf, we can find x1 such that G(x) > 0 for all x >= x1. ---> Since lim F'(x)/G'(x) exists, we can find x2 >= x1 such that G'(x) <> 0 for all x >= x2. ---> Since lim F'(x)/G'(x) = 0, we can find x3 >= x2 such that (*) -eps/2 < F'(x)/G'(x) < eps/2 for all x >= x4. (We are saving half of eps for further restrictions). On the interval [x3, inf), G' is defined everywhere, hence it has the Intermediate Value Property; since it is never zero, it cannot change sign. Since G(x) -> inf, G'(x) > 0 on [x3, inf). Clear the fractions in the inequalities (*) and simplify: eps * G'(x) + 2F'(x) > 0 and eps * G'(x) - 2F'(x) > 0 so that eps * G(x) + 2F(x) and eps * G(x) - 2F(x) are increasing on [x3, inf). Comparing values at x3 and at x > x3, we obtain eps * G(x) + 2F(x) > eps * G(x3) + 2F(x3) and eps * G(x) - 2F(x) > eps * G(x3) - 2F(x3) for all x > x3. Divide by the positive quantity 2G(x) and isolate F(x)/G(x): eps eps * G(x3) + 2F(x3) F(x) eps eps * G(x3) - 2F(x3) - --- + -------------------- < ---- < --- - -------------------- 2 2G(x) G(x) 2 2G(x) ----> The fractions const/G(x) above converge to 0 since G -> inf, so we can find x4 >= x3 such that for every x >= x4, the magnitudes of both of these fractions are less than eps/2. That was the other eps/2 we used up. Hence -eps < F(x)/G(x) < eps for every x > x4. To adhere to our notation, we declare x0 = x4. This proved that lim F(x)/G(x) = 0 as x -> inf. Finite limits other than 0: -------------------------- Suppose g(x) -> inf and f'(x) / g'(x) -> L as x -> inf. Then f(x)/g(x) -> L. Proof. Define F(x) = f(x) - Lg(x) and G(x) = g(x), and apply the special case result: f'(x) - Lg'(x) f(x) - Lg(x) -------------- -> 0, so ------------ -> 0, and simplify. g'(x) g(x) One-sided limit at c+: ---------------------- Change the variable by x = c + 1/t where t -> inf. The effect of Chain Rule is that both numerator and denominator are multiplied by (-1/t^2), and that cancels out. Similar changes of variable can establish the Rule for x -> c- and for x -> -inf. Infinite limits: ---------------- We want to solve F(x)/G(x) > K (K given) for x sufficiently large. We start as in the case lim = 0, but then change conditions. First, we make F'(x)/G'(x) > (K+1), so that F(x) - (K+1)G(x) increases, so F(x) (K+1)G(x3) - F(x3) ---- > K + 1 - ------------------ G(x) G(x) and the rightmost fraction can be made less than 1 as long as x >= x4 >= x3. Finished. ============================================================================== From: acs@cheerful.com Subject: Re: L'Hospital Rule when f(x) => 00 Date: Thu, 02 Dec 1999 14:33:10 GMT Newsgroups: sci.math In article <81v43k$21o$1@bgtnsc01.worldnet.att.net>, "Nathaniel Silver" wrote: > acs@cheerful.com wrote: > >Can someone give a simple proof of L'Hospital Rule for the case > >when f(x) and g(x) => 00? That is, let f and g be functions from > >R into R such that f(x) and g(x) => 00 when x=> 00 and that > >lim f'(x)/g'x)=L > >x=>00 > >Then, lim f(x)/g(x) = L, too. > >x=>00 > >I gave a proof for this rule, based on Cauchy's > Mean Value Theorem, but it's really complicated. > > Here is a plausibility argument, which goes through > for nice, nice cases. We "prove" a preliminary lemma > that has the flavor of the argument but can be assumed. > > Lemma: L'Hopital's Rule. > Suppose f /g and f and g never have properties > that disable the proof of this theorem. > Let f(a) = 0 = g(a) and > suppose lim (f'(x)/g'(x)) = L. > x=>a > Then lim (f(x)/g(x)) = L. > x=>a > > "Proof:" > > lim (f(x)/g(x)) > x=>a > > = lim [(f(x)-f(a))/(g(x)-g(a))] > x=>a > > = lim [(f(x)-f(a))/(x-a) / (g(x)-g(a))/(x-a)] > x=>a > > = lim [(f(x)-f(a))/(x-a)] / lim[(g(x)-g(a))/(x-a)] > x=>a x=>a > > = f'(a) / g'(a) = L Actually, you don't have to assume f'(a) and g'(a) exist and the second is diferent from zero. All you have to assume is the existence of lim (x=>a) f'(x)/g'(x) exists and that f' and g' exist on a neighborhood of a. Proof: Let f and g be functions continuous on the closed interval [a; b] and differentiable on (a; b) such that f(a) = 0 and g(a) = 0. In addition, suppose lim x=>a f'(x)/g'(x) = L. Then, for any eps >0, there's a 0 < delta < b-a such that if a < x < a+ delta then |f'(x)/g'(x) - L| < eps (inequality 1) As f and g are continuous on [a; b] and differentiable on (a; b) , then such conditions hold for [a; x] and (a; x)if x is in (a; b) . So, according to Cauchy's Mean Value Theorem, there exists c belonging to (a; x) (c depends on x) such that [f(x) - f(a)]/[g(x)-g(a)] = f(x)/g(x) (because f(a) = g(a) = 0) = f'(c)/g'(c). As a < c < x 0, there exists delta > 0 such that , if a a+ f(x)/g(x) = L. In a similar way, we can prove that lim x=> a- f(x)/g(x) = L, if a is a interior point of the domain of f/g. But this proof doesn't hold if x=> inf. If f(x) and g(x) => 0 when x=> inf, a simple proof can be given just making t =1/x and applying L'Hospital Rule, 0/0 form, to the case when t=>0+. But things get much more difficult when x=> inf and so do f(x) and g(x). Artur Sent via Deja.com http://www.deja.com/ Before you buy.