LHopitals Rule in Chemistry: Irreversible Morphing into Reversible
31-Mar-2009 L'Hôpital's Rule is discussed in the case of a reversible isothermal expansion with the idea of reinforcing ideas from elementary calculus. We ...
On LHôpitals Rule 1 The three theorems
The extended form also applies to forms of the type ?/? and to limits as x ? ??. 1 The three theorems. Theorem 1 (Baby L'Hôpital's Rule) Let f(x) and g(x)
31 LHopitals Rule
31 L'Hopital's Rule. 31.1 Limit of indeterminate type. Some limits for which the substitution rule does not apply can be found by using inspection.
LHopitals Rule and Expansion of Functions in Power Series
In a similar way further coefficients may be found. By stopping at any de- sired place as for example
31. LHopitals Rule
31. L'Hopital's Rule. 31.1. Limit of indeterminate type. Some limits for which the substitution rule does not apply can be found by using inspection.
Differentials and Higher Ordered Approximations
L'Hopital's Rule. Derivatives are defined using limits. Here we have an application where derivatives are used to find limits. L'Hopital's Rule for Limit at
Indeterminate Forms and LHopitals Rule When we computed limits
form of the third limit is called an indeterminate form; basically we don't know what an answer of this form means. L'Hopital's Rule
Indeterminate Forms and LHopitals Rule When we computed limits
form of the third limit is called an indeterminate form; basically we don't know what an answer of this form means. L'Hopital's Rule
Section 4.7: LHopitals Rule Growth
https://www.math.arizona.edu/~mgilbert//Math_122B/Lecture_Notes/Section_4.7_Lecture_Notes_122B.pdf
7.7 Indeterminate Forms and LHospitals Rule
If l'Hospital's Rule doesn't apply explain why. 1. 2. 3. 4. 5. 6.
[PDF] 31 LHopitals Rule
31 L'Hopital's Rule 31 1 Limit of indeterminate type Some limits for which the substitution rule does not apply can be found by using inspection
[PDF] LHôpitals Rule and Indeterminate Forms - Arizona Math
L'Hôpital's Rule allows us to evaluate these kinds of limits without much effort It also allows us to deal with different indeterminate forms
[PDF] 312 LHôpitals rule - Karlinmffcunicz
The verification of l'Hôpital's rule (omitted) depends on the mean value theorem 31 2 1 Example Find lim x?0 x2 sin x
[PDF] 77 Indeterminate Forms and LHospitals Rule
The use of l'Hospital's Rule is indicated by an H above the equal sign: H= 1 lim x?2 x ? 2
[PDF] LHôpitals rule practice problems 21-121 - CMU Math
L'Hôpital's rule practice problems 21-121: Integration and Differential Equations Find the following limits You may use L'Hôpital's rule where
[PDF] Indeterminate Forms and LHospitals Rule
THEOREM (L'Hospital's Rule): Suppose f and g are differentiable and g?(x) = 0 near a (except possibly at a) Suppose that lim x?a f(x) = 0 and lim
[PDF] 8 LHOPITALS RULE
L'Hopital's Rule says that the limit of an indeterminant quotient of functions should be the same as the limit of of the quotient of the derivatives of those
[PDF] LHôpitals Rule
Applying L'Hopital would give a wrong answer! To make f(x) continuous we set f(0) = limx?0 f(x) = e?? = 0
[PDF] 1 LHospitals Rule - CUHK Mathematics
1 L'Hospital's Rule Another useful application of mean value theorems is L'Hospital's Rule It helps us to evaluate limits of “indeterminate forms” such as
[PDF] On LHôpitals Rule 1 The three theorems
Theorem 1 (Baby L'Hôpital's Rule) Let f(x) and g(x) be continuous functions on an interval containing x = a with f(0) = g(0) = 0 Suppose
What is an example of L Hôpital's rule?
Example application of l'Hôpital's rule to f(x) = sin(x) and g(x) = ?0.5x: the function h(x) = f(x)/g(x) is undefined at x = 0, but can be completed to a continuous function on all of R by defining h(0) = f?(0)/g?(0) = ?2.What is the simple explanation of L Hopital's rule?
Formally, L'Hopital's rule says that if you have some functions, like f(x) and g(x), and both of them approach zero as x goes to some number, like C, then the limit as x approaches C of the ratio of these functions is equal to the limit as x approaches C of the ratio of the derivatives of these functions.- The L'Hospital rule failed because limx??(sin3x)? does not exist.
On L"Hˆopital"s Rule
There are three versions of L"Hˆopital"s Rule, which I call "baby L"Hˆopital"s rule", "macho L"Hˆopital"s rule" and "extended L"Hˆopital"srule". The baby and macho versions refer to the problem of evaluating lim xaf(x)/g(x), where lim xaf(x) = limxag(x) = 0. In other words, indeterminate forms of the type "0/0", withafinite. (Also to limits asxa+and asxa.) The extended form also applies to forms of the type/and to limits as x .1 The three theorems
Theorem 1 (Baby L"Hˆopital"s Rule)Letf(x)andg(x)be continuous functions on an interval containingx=a, withf(0) =g(0) = 0. Suppose thatfandgare differentiable, and thatfandgare continuous. Finally, suppose thatg(a)= 0. Then lim xaf(x) g(x)= limxaf (x)g(x)=f(a)g(a). Also, lim xa+f(x) g(x)= lim xa+f (x)g(x) and lim xa-f(x) g(x)= lim xa-f (x)g(x). The baby version is easy to prove, and is good enough to compute limits like lim x0sin(2x) x+x2.(1) However, it isn"t good enough to compute limits like lim x01cos(2x) x2,(2) since in that caseg(0) = 0. To solve problems like (2), we need the macho version: 1 Theorem 2 (Macho L"Hˆopital"s Rule)Suppose thatfandgare contin- uous on a closed interval[a,b], and are differentialble on the open interval (a,b). Suppose thatg(x)is never zero on(a,b), and thatlimxa+f(x)/g(x) exists, and thatlimxa+f(x) = limxa+g(x) = 0. Then lim xa+f(x) g(x)= lim xa+f (x)g(x). Note that this theorem doesn"t require anything aboutg(a), just about howgbehaves to the right ofa. An analogous theorem applies to the limit asxa(and requiresfandgandfandgto be defined on an interval thatendsata, rather than one that starts ata). You can combine the two to get a theorem about an overall limit asxa. The conclusion of Macho L"Hˆopital"s Rule relates one limit (off/g) to another limit (off/g), and not to the value off(a)/g(a). This is what allows the theorem to be used recursively to solve problems like (2). Finally, we have the Theorem 3 (Extended L"Hˆopital"s Rule)L"Hˆopital"s rule applies to in- definite forms of type "/" as well as "0/0", and applies to limits as x as well as to limitsxa. In all of these cases, lim f(x) g(x)= limf(x)g(x).2 Proofs of the baby and macho theorems
Suppose thatf(a) =g(a) = 0 andg(a)= 0. Then, for anyx,f(x) = f(x)f(a) andg(x) =g(x)g(a). But then, lim xaf(x) g(x)= limxaf(x)f(a)g(x)g(a) = lim xa[f(x)f(a)]/(xa) [g(x)g(a)]/(xa) limxa([f(x)f(a)]/(xa)) limxa([g(x)g(a)]/(xa)) f(a) g(a), 2 since, by definition,f(a) = limxaf(x)f(a)xaandg(a) = limxag(x)g(a)xa. Sincefandgare assumed to be continuous, this is also lim xaf(x) limxag(x)= limxaf (x)g(x).That proves the baby version.
To prove the macho version, we first need a lemma: Theorem 4 (Souped up Mean Value Theorem)Iff(x)andg(x)are continuous on a closed interval[a,b]and differentiable on the open interval (a,b), then there is a pointc, betweenaandb, where (f(b)f(a))g(c) = (g(b)g(a))f(c).(3) (Wheng(x) =x, this is the same as the usual MVT.)Proof of Souped up MVT: Consider the function
h(x) = (f(x)f(a))(g(b)g(a))(f(b)f(a))(g(x)g(a)). This is continuous on [a,b] and differentiable on (a,b), with h (x) =f(x)(g(b)g(a))g(x)(f(b)f(a)). Note thath(a) = 0 =h(b). By Rolle"s Theorem, there a spotcwhere h (c) = 0. Buth(c) = 0 is the same as equation (3). Proof of Macho L"Hˆopital"s Rule: By assumption,fandgare differ- entiable to the right ofa, and the limits offandgasxa+are zero. Definef(a) to be zero, and likewise defineg(a) = 0. Since these values agree with the limits,fandgare continuous on some half-open interval [a,b) and differentiable on (a,b). For anyx(a,b), we have thatfandgare differentiable on (a,x) and continuous on [a,x]. By the Souped up MVT, there is a pointcbetweenaand xsuch thatf(c)g(x) =f(x)g(c). In other words,f(c)/g(c) =f(x)/g(x). Also, asxapproachesa,calso approachesa, sincecis somewhere between xanda. But then lim xa+f(x) g(x)= lim xa+f (c)g(c)= lim ca+f (c)g(c).That last expression is the same as lim
xa+f(x)/g(x). 33 Proving the extended theoremWe"re going to use a single trick, over and over again. Namely,we can always
rewritexas 1/(1/x),f(x) as 1/(1/f(x)) andg(x) as 1/(1/g(x)).SupposeL= limxaf(x)
g(x), where bothfandggo to(or) asxa.Also suppose thatLis neither 0 nor infinite. Then
L= limxaf(x)
g(x)= limxa1/g(x)1/f(x). Since 1/g(x) and 1/f(x) go to zero asxa, we can apply the (baby or macho) L"Hˆopital"s rule to this limit:L= limxa(1/g)
(1/f) = lim xag(x)/g(x)2 f(x)/f(x)2 = lim xaf(x)2g(x) g(x)2f(x) = lim xaf(x)2 g(x)2limxag (x)f(x) L2 limxa[f(x)/g(x)]. SinceL=L2/limxa[f(x)/g(x)],Lmust equal limxa[f(x)/g(x)], which is what we wanted to prove. This argument only works for finite and nonzero values ofL. However, if L= 0, we can apply the same argument to the limit of (f(x) +g(x))/g(x), which then does not equal zero. The upshot is that1 + lim
xaf(x) g(x)= limxaf(x) +g(x)g(x)= limxaf (x) +g(x)g(x)= 1 + limxaf (x)g(x), hence that lim(f/g) = lim(f/g). Finally, if lim(f/g) =, look instead at lim(g/f), which is then zero, so the previous reasoning applies. Since0 = lim(g/f) = lim(g/f), lim(f/g) must be infinite. By the Souped up
MVT,f/ghas the same sign asf/g, so we must have lim(f/g) = lim(f/g). Now that we have L"Hˆopital"s Rule for limits asxa(orxa+or xa), we consider what happens asx . Define a new variable 4 t= 1/x, so thatx is the same ast0+. Then lim xf(x) g(x)= lim t0+f(1/t)g(1/t). But we know how to apply L"Hˆopital"s Rule to limits ast0, so this turns into lim t0+d dtf(1/t) d dtg(1/t)= lim t0+f(1/t)/t2g(1/t)/t2= lim t0+f (1/t)g(1/t).Converting back tox= 1/t, we get
lim xf (x) g(x), which is what we wanted. Computing a limit asx is similar, only witht0instead oft0+. That completes the proof of the Extended L"Hˆopital"s Rule. 5quotesdbs_dbs22.pdfusesText_28[PDF] l'immigration aux etats unis aujourd'hui
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