7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

Transcription

1 7 Improper Integrls, Exp, Log, Arcsin, nd the Integrl Test for Series We hve now ttined good level of understnding of integrtion of nice functions f over closed intervls [, b]. In prctice one often wnts to extend the domin of integrtion nd consider unbounded intervls such s [, ) nd (, b]. The simplest non-trivil exmples re the infinite trumpets defined by the res under the grphs of x t for t > 0, i.e., the improper integrls A t = x t dx. We will see below tht A t hs well defined mening if t >, but becomes unbounded for t. One is lso interested in integrls of functions f over finite intervls with f being unbounded. The nturl exmples re given (for t > 0) by B t = 0 x t dx. Here it turns out tht B t is well defined, i.e., hs finite vlue, if nd only if t <. In prticulr, neither A nor B mkes sense. Moreover, A t + B t = 0 x t dx is unbounded for every t > Improper Integrls Let f be function defined on the interior of possibly infinite intervl J such tht either its upper endpoint cll it b, is or f becomes unbounded s one pproches b. But suppose tht the lower endpoint cll it, is finite nd tht f() is defined. In the former cse the intervl is unbounded, while in the ltter cse the intervl is bounded, but the function is unbounded. We

2 will sy tht the integrl of f over J exists iff the following two conditions hold: (7..) (i) For every u (, b), f is integrble on [, u]; nd (ii) the limit exists. u lim f(x)dx u b, u<b When this limit exists, we will cll it the integrl of f over J nd write it symboliclly s (7..2) f(x)dx. Sometimes we will lso sy tht the integrl (7..2) converges when it mkes sense. Similrly, if is either or is finite point where f becomes unbounded, but with b finite point where f is defined, one sets (7..3) f(x)dx = lim f(x)dx u, u> u when the limit on the right mkes sense. Lemm 7. We hve f(x)dx = c f(x)dx + f(x)dx c (, ) c nd f(x)dx = c f(x)dx + f(x)dx c (, b), whenever the integrls mke sense. c 2

3 Proof. We will prove the first dditivity formul nd leve the other s n exercise for the reder. Pick ny c (, ). or ll rel numbers u > c, we hve by the usul dditivity formul, u f(x)dx = c f(x)dx + u f(x)dx. c Thus which equls f(x)dx = lim u c f(x)dx + u c f(x)dx, c u f(x)dx + lim u c f(x)dx = c f(x)dx + c f(x)dx. If J is n intervl with both of its endpoints being problemtic, we will choose point c in (, b) nd put (7..4) f(x)dx = c f(x)dx + f(x)dx c if both the improper integrls on the right mke sense. One cn check using Lemm 7. bove tht this definition is independent of the choice of c. When n improper integrl does not mke sense, we will cll it divergent. Otherwise it is convergent. Proposition Let t be positive rel number. Then for t >, (A) x dx = t t, with the improper integrl on the left being divergent for t. 3

4 On the other hnd, if t (0, ), (B) 0 x dx = t t, with the improper integrl on the left being divergent for t. Proof. ll t : Hence for t >, We lernt in section 5.4 tht for ll, b R with < b, nd for lim u u dx = lim xt u x dx = b t t. t t n= ( t)u + t t = t, becuse the term goes to zero s u goes to. If t <, this term goes u t to s u goes to, nd so the integrl is divergent. Finlly let t =. Then we cnnot use the bove formul. But for ech N, we hve the inequlity N N n x dx. The reson is tht the sum on the left is lower Riemnn sum for the function f(x) = over the intervl [, N] reltive to the prtition P : < 2 <... < x N. So, if the improper integrl of this function exists over [, ), the infinite series N n = lim N n n= must converge. But we hve seen in Chpter 2 tht this series diverges. So the integrl is divergent for t =, nd (A) is proved in ll cses. The proof of (B) is very similr nd will be left s n exercise. n= 4

5 7.2 Inverse functions Suppose f is function with domin X nd imge (or rnge) Y. By definition, given ny x in X, there is unique y in Y such tht f(x) = y. But this definition of function is not eglitrin, becuse it does not require tht unique number x be sent to y by f; so y is specil but x is not. A relly nice kind of function is wht one clls one-to-one (or injective) function. By definition, f is such function iff (7.2.) f(x) = f(x ) = x = x. In such cse, we cn define n inverse function g with domin Y nd rnge X, given by (7.2.2) g(y) = x iff f(x) = y. Clerly, when such n inverse function g exists, one hs (7.2.3) g f = X nd f g = Y, where X, resp. Y, denotes the identity function on X, resp. Y. We will be concerned in this chpter with X, Y which re subsets of the rel numbers. Proposition 2 Let f be one-to-one function with domin X R nd rnge Y, with inverse g. Suppose in ddition tht f is differentible t x with f (x) 0. Then g is differentible t y = f(x) nd we hve for ll x in X with y = f(x). g (y) = f (x) Note tht if we know priori tht f nd g re both differentible, then this is esy to prove. Indeed, in tht cse their composite function g f, which is the identity on X, would be differentible. By differentiting the identity g(f(x)) = x 5

6 with respect to x, nd pplying the chin rule, we get g (f (x)) f (x) =, becuse the derivtive of x is. Done. Proof. We hve to compute the limit g(y + h) g(y) L = lim. h 0 h Since f is differentible, it is in prticulr continuous, which implies tht So, if we set lim f(x + h 0 h ) = f(x). h = f(x + h ) f(x), then h 0 when h 0. Writing y = f(x), we then get, fter pplying g to y + h = f(x + h ), g(y + h) = x + h, or in other words, h = g(y + h) g(y). We clim tht, since f is one-to-one, we must lso hve h 0 when h 0. Suppose not. Then for some ε > 0, h = g(y + h) g(y) is ε for ll h close to 0. But this will led to f sending two distinct numbers, with x + h being one of them, to the sme number y + h; the other number will be close to x, of distnce less thn ε. This contrdicts the fct tht f is one-to-one. Hence the Clim. Hence L = lim h 0 h f(x + h ) f(x), which is the inverse of f (x). It mkes sense becuse f (x) is by ssumption non-zero. Note tht this proof shows tht g is not differentible t ny point y = f(x) if f is zero t x. 6

7 It is helpful to note tht mny function is not one-to-one in its mximl domin, but becomes one when restricted to smller domin. To give simple exmple, the squring function f(x) = x 2 is defined everywhere on R. But it is not one-to-one, becuse f() = f( ). However, if we restrict to the subset R + of non-negtive rel numbers, f is one-to-one nd so we my define its inverse to be the squre-root function g(y) = y, y R +. Another exmple is provided by the sine function, which is periodic of period 2π nd hence not one-to-one on R. But it becomes one when restricted to [ π/2, π/2]. 7.3 The nturl logrithm For ny x > 0, its nturl logrithm is defined by the definite integrl (7.3.) log x = x dt t. Some write ln(x) insted, nd some others write log e x. When 0 < x <, this signifies the negtive of integrl of from x to. Consequently, log x is t positive if x >, negtive if x < nd equls 0 t x =. Proposition 3 () log = 0. (b) log x is differentible everywhere in its domin R + = (0, ) with derivtive x. (c) (ddition theorem) For ll x, y > 0, log(xy) = log x + log y. (d) log x is strictly incresing function. 7

8 (e) log x becomes unbounded in the positive direction when x goes to nd it is unbounded in the negtive direction when x goes to 0. (f) log x is integrble on ny finite subintervl of R +, nd its indefinite integrl is given by log x dx = x log x x + C. (g) log x goes to (resp. ) slowly when x goes to 0 (resp. ); more precisely, (i) nd (ii) lim x log x = 0 x 0 log x lim x x = 0. (h) The improper integrl of log x over (0, b] exists for ny b > 0, with log x dx = b log b b. 0 Property (c) is very importnt, becuse it cn be used to trnsform multiplictive problems into dditive ones. This ws the motivtion for their introduction by Npier in 66. Property (b) is lso importnt. Indeed, if we ssume only the properties (),(c),(d),(e) for function f on R +, there re lots of functions (logrithms) which stisfy these properties. But the sitution becomes rigidified with unique solution once one requires (b) s well. This is why log is clled the nturl logrithm. The other (unnturl) choices will be introduced towrds the end of the next section. Proof. (): This is immedite from the definition. (b): For ny x, the function is continuous on [0, x], hence by the First t Fundmentl Theorem of Clculus, log x is differentible with derivtive /x. (c): Fix ny y > 0 nd consider the function of x defined by l(x) = log(xy) on {x > 0}. 8

9 Since it is the composite of the differentible functions x xy nd u log u, l is lso differentible. Applying the chin rule, we get ( ) l (x) = y = yx x. Thus l nd log both hve the sme derivtive, nd so their difference must be independent of x. Write l(x) = log x + c. Evluting t x =, nd noting tht log = 0 nd l() = log y, we see tht c must be log y, proving the ddition formul. (d): As we sw in Chpter 4, differentible function f is strictly incresing iff its derivtive is positive everywhere. When f(x) = log x, the derivtive, s we sw bove, is /x, which is positive for x > 0. This proves (d). (e): As log x is strictly incresing nd since it vnishes t, its vlue t ny number x 0 >, for instnce t x 0 = 2, is positive. By the ddition theorem nd induction, we see tht for ny positive integer n, (7.3.2) log(x n 0) = n log x 0. Consequently, s n goes to, log(x n 0) goes to s well. This proves tht log x is unbounded in the positive direction. For the negtive direction, note tht for ny positive x <, log x is negtive (since log = 0 nd log x is incresing). Applying (8,2,3) with x 0 replced by x, we deduce tht log(x n ) goes to s n goes to. Done. (f) Since log x is continuous, it is integrble on ny finite intervl in (0, ). Moreover, It derivtive exists nd is continuous t ll x > 0. So we my pply integrtion by prts to f(x) = log x nd g(x) = x (so tht g (x) = ) to obtin log x dx = x log x x d (log x)dx. dx The ssertion (f) now follows since the expression on the right is x log x x + C. (g): Put L = lim x log x x 0 nd u = x, f(u) = log ( u ), nd g(u) = u. 9

10 Then we hve L = lim u f(u) g(u). Then by (e), f(u) nd g(u) pproch s u goes to, nd both these functions re differentible t ny positive u. (All one needs is tht they re differentible for lrge u.) Since u (x) = = u 2 nd f(u) = log x, we x 2 hve by the chin rule, Since g (u) = for ll u, we then get f (u) = df/dx du/dx = /x u 2 = u. lim u f (u) g (u) = 0. So we my pply L Hopitl s rule (see the Appendix to this chpter), nd conclude tht f (u) L = lim u g (u) = 0, giving (i). The proof of (ii) is similr nd will be left for the reder to check. (h): The improper integrl of log x exists over (0, b] iff the following limit exists: L = lim log t dt. x 0 x Thnks to (f), we hve log t dt = (t log t t) b x = (b log b b) x log x + x. x To prove (h) we need to show tht lim (x log x x) = 0, x 0 which is consequence of (g). 0

11 Remrk on infinite products: Suppose x, x 2,..., x n,... is n infinite sequence of positive rel numbers. Often we will wnt to consider infinite products of the form P := x n, n= which my not in generl converge to finite rel number. Since the logrithm function converts multipliction into ddition, it is convenient to consider the log of this product P, nd consider the infinite series log P := log(x n ). n= We will sy tht P converges when log P does. 7.4 The exponentil function In view of the discussion in section 7.2, nd the fct (see prts (d), (e) of Proposition 3) tht log x is strictly incresing function with domin R + nd rnge R, we cn define the exponentil function, exp(x) for short, to be the inverse function of log x. Note tht exp(x) hs domin R nd rnge R +.

12 Proposition 4 () exp(0) =. (b) exp(x) is differentible everywhere in R with derivtive exp(x). (c) (ddition theorem) For ll x, y R, exp(x + y) = exp(x) exp(y). (d) exp(x) is strictly incresing function. (e) exp(x) becomes unbounded in the positive direction when x goes to nd it goes to 0 when x goes to. (f) exp(x) is integrble on ny finite subintervl of R +, nd its indefinite integrl is given by exp(x) dx = exp(x) + C. (g) As x goes to (resp. ), exp(x) goes to (resp. 0) fster thn ny polynomil p(x) goes to (resp. ), i.e., (i) lim x p(x) exp(x) = 0 nd (ii) lim p(x) exp(x) = 0. x (h) The improper integrl of exp(x) over (, b] exists for ny b > 0, with exp(x) dx = exp(b). Proof. (): Since log = 0, nd s exp log is the identity function, we see tht exp(0) = exp(log ) =. 2

13 (b): Note tht the derivtive x of log x is nowhere zero on its domin R +. So we my pply Proposition 2 with f = log, g = exp nd y = log x to get the everywhere differentibility of exp, with (7.4.) d dy exp(y) = = (log x) /x = x = exp(y). (c): Fix x, y in R. Then we cn find (unique) u, v in R + such tht x = log u nd y = log v. Applying prt (c) of Proposition 3, we then obtin exp(x + y) = exp(log u + log v) = exp(log(uv)) = uv. The ssertion follows becuse u = exp(x) nd v = exp(y). (d): Let x > y be rbitrry in R. Write x = log u, y = log v s bove. Since log is strictly incresing, we need u > v. Since u = exp(x) nd v = exp(v), we re done. (e): Suppose exp(x) is bounded from bove s x goes to infinity, i.e., suppose there is positive number M such tht exp(x) < M for ll x > 0. Then, since log is strictly incresing function, x = log(exp(x)) would be less thn log(m) for ll positive x, which is bsurd. So exp(x) must be unbounded s x becomes lrge. To show tht exp(x) pproches 0 s x goes to, we need to show tht for ny ε > 0, there exists T < 0 such tht (7.4.2) exp(x) < ε whenever x < T. Applying the logrithm to both inequlities, writing T = log u for unique u (0, ) nd ε = exp(ε), nd using the fct tht log is one-to-one function, we see tht (7.4.2) is equivlent to the sttement (7.4.3) x < ε whenever log x < log u, which evidently holds by the properties of the logrithm. Hence exp(x) goes to 0 s x goes to. (f) Since exp(x) is continuous, it is integrble on ny finite subintervl of R. In fct, since exp(x) is its own derivtive, it is its own primitive s well, proving the ssertion. 3

14 (g): To prove (i), it suffices to show tht for ny j 0, the limit L = lim x x j exp(x) exists nd equls 0. When j = 0 this follows from (e), while the ssertion for j > follows from the cse j =, which we will ssume to be the cse from here on. It then suffices to show tht the function exp(x)/x goes to s x goes to. By tking logrithms, this becomes equivlent to the sttement tht x/ log x goes to s log x, nd hence x, goes to, which is wht we proved in our proof of the limit (i) of Proposition 3, prt (g). (One cn lso pply, with cre, L Hopitl s rule directly to the quotient x/ exp(x) nd thereby estblish (i).) The proof of (ii) is similr nd will be left for the reder to check. (h): The improper integrl of exp(x) exists over (, b] iff the following limit exists: Thnks to (f), we hve L = lim x x exp(t)dt. exp(t) dt = exp(b) exp(x). To prove (h) we need only show tht which is consequence of (e). x lim exp(x) = 0, x How mny differentible functions re there which re derivtives (nd hence primitives) of themselves? This is nswered by the following Lemm 7.2 Let f be ny differentible function on n open intervl (, b), possibly of infinite length, stisfying f (x) = f(x) for ll x in (, b). Then there exists sclr c such tht f(x) = c exp(x) x (, b). Moreover, if f(0) =, then f(x) equls exp(x) on this intervl. 4

15 Proof. Define function h on (, b) by h(x) = f(x) exp(x), which mkes sense becuse exp never vnishes nywhere. Since f nd exp re differentible, so is h, nd by the quotient rule we hve h (x) = f (x) exp(x) f(x) exp (x) exp(x) 2 = f(x) exp(x) f(x) exp(x) exp(x) 2 = 0, becuse f nd exp re derivtives of themselves. Therefore h(x) must be constnt c, sy. The Lemm now follows. Definition Put e = exp(). Equivlently, e is the unique, positive rel number whose nturl logrithm is. It is not hrd to see tht, to first pproximtion, 2 < e < 4. One cn do much better with some work, of course, nd lso show tht e is irrtionl, even trnscendentl. A nturl question now rises. We hve worked with the power function x before, for ny positive rel number, which stisfies the sme ddition rule s exp x, i.e., (7.4.4) x+y = x y.. Wht is the reltionship between e x nd exp(x). The nswer is very stisfying. Proposition 5 For ll x in R, we hve exp(x) = e x. We will need the following: Lemm 7.3 log(e x ) = x, for ll x R. 5

16 Proof. This is cler for x n integer m. Since we hve for ny n > 0, = log(e) = log((e /n ) n ) = n log(e /n ), we deduce tht log(e /n ) = /n. Consequently, log(e m/n ) = m, m/n Q. n For ny rel number x, e x is defined to be the limit lim αn x e αn, where α n is sequence of rtionl numbers α n coverging to x. Since u log(u) is continuous function, we get log( lim α n x eα n ) = lim α n x log(eα n ) = lim α n = x. α n x Proof of Proposition 5. Applying Lemm 7.3, nd using the fct tht exp(x) is the inverse function of logrithm, we get log(exp(x)) = x = log(e x ), x R. Since log is one-to-one, we must hve exp(x) = e x. Just like exp(x), the power function x is, for ny positive number, strictly incresing function. One clls the inverse of x by the nme logrithm to the bse nd denotes it by log y. Then log stisfies mny of the properties of log, but its derivtive is not /x (if e). Proposition 6 Fix > 0. Then nd d dx (x ) = x log, d dy (log y) = y log. Proof. Note tht x = exp(log( x )) = e x log, 6

17 which is the composite of x x log nd u e u. Applying the chin rule, we obtin d dx (x ) = e x log log = x log. The formul for the derivtive of log y follows from this nd Proposition 2. The bsic hyperbolic functions re defined s follows: (7.4.5) sinh x = ex e x, cosh x = ex + e x, tnh(x) = sinh x 2 2 cosh x. The other hyperbolic functions re given by the inverses of these. 7.5 rcsin, rccos, rctn, et l The sine function, which is defined nd differentible everywhere on R, is periodic with period 2π nd is therefore not one-to-one function. It however becomes injective when the domin is restricted to [ π/2, π/2]. One clls the corresponding inverse function the rcsine function, denoted rcsin x. Clerly, the domin of rcsin x is [, ] nd the rnge is [ π/2, π/2]. Moreover, since the derivtive cos x of sin x is non-zero, in fct positive, in ( π/2, π/2), we my pply Proposition 2 nd deduce tht the rcsine function is differentible on ( π/2, π/2), with (7.5.) d (rcsin y) = dy cos x = sin 2 x = y, 2 where y = sin x. One notes similrly tht cos x, resp. tn x, is one-to-one on [0, π], resp. ( π/2, π/2), nd defines its inverse function rccos x, resp. rctn x, with domin [, ] nd rnge [0, π]. Arguing s bove, we see tht rccos x is differentible on (0, π) nd rctn x is differentible on ( π/2, π/2), with (7.5.2) d dy (rccos x) = y 2 nd (7.5.3) d dy (rctn x) = + y 2 7

18 Consequently, we obtin dy (7.5.4) + y = rctn y + C 2 nd (7.5.5) dy y 2 = rcsin y + C. This incidentlly brings to close our quest to integrte rbitrry rtionl functions, which we begn in the previous chpter, where we reduced the problem to the evlution of the integrl on the left of (7.5.4). It should be notes tht it is mircle tht we cn evlute the reciprocl of the squre root of y 2 (for y ) in terms of rcsin y. If one tried to integrte for polynomil f of degree n > 2, the problem becomes f(y) forbiddingly difficult. Even for n = 3, one needs to use elliptic functions. 7.6 A useful substitution A very useful substitution to del with trigonometric integrls is to set (7.6.) u = tn(x/2), which implies tht x = 2 rctn u. Note tht du (7.6.2) dx = 2 sec2 (x/2) 2 ( + tn2 (x/2)) = + u2, 2 (7.6.3) dx = 2 + u 2 du, (7.6.4) sin x = 2 sin(x/2) cos(x/2) = 2 tn(x/2) sec 2 (x/2) = 2u + u 2, 8

19 nd since cos 2 (x/2) + sin 2 (x/2) =, (7.6.5) cos x = cos2 (x/2) sin 2 (x/2) cos 2 (x/2) + sin 2 (x/2) = tn2 (x/2) + tn 2 (x/2) = u2 + u 2. For exmple, suppose we hve to integrte dx I = sin x. Using the substitution bove, which is justifible here, we get 2 I = 2u/( + u 2 ) + u du = du 2 2 2u + u. 2 Since we get I = 2u + u = 2 ( u) = d ( ), 2 du u ( ) d u = u + C = tn(x/2) + C. 7.7 The integrl test for infinite series When we discussed the question of convergence of infinite series in chpter 2, we gve vrious tests one could use for this purpose, t lest for series with non-negtive coefficients. Here is nother test, which cn t times be helpful. Proposition 7 Consider n infinite series S = n, whose coefficients stisfy n= n = f(n), for some non-negtive, monotone decresing function f on the infinite intervl [, ). Then S converges iff the improper integrl I = f(x)dx converges. 9

20 Proof. For ny integer N >, consider the prtition P N : < 2 <... < N of the closed intervl [0, N]. Then, since f is monotone decresing, the upper nd lower sums re given by U(f, P N ) = N nd L(f, P N ) = N + N. Suppose f is integrble over [, ). Then it is integrble over [, N] nd N N f(x)dx N 2 + N. As N goes to infinity, this gives S f(x)dx, which implies tht S is convergent. To prove the converse we need to be bit more wily. Suppose S converges. Note tht f is integrble over [, ) iff the series T = n= b n converges, where b n = n+ f(x)dx n But since f is monotone decresing over ech intervl [n, n + ], the re under the grph of f is bounded bove (resp. below) by the re under the constnt function x f(n) = n (resp. x f(n + ) = n+ ). Thus we hve, for every n, n+ b n n. 20

21 Summing from n = to nd using the comprison test (see chpter 2), we get S T S. Thus T converges s well. As consequence we see tht for ny positive rel number t, the series S t = n t converges iff the improper integrl I t = n= x t dx converges. We hve lredy seen tht I t is convergent iff t >. So the sme holds for S t. But recll tht in the specil cse t =,we deduced the divergence of I from tht of S. Appendix: L Hôpitl s Rule It ppers tht the most populr mthemticin for Clculus students is Mrquis de L Hôpitl, who ws prolific during the end of the seventeenth century. Everyone likes to use his rule, but two things must be tken due note of. The first is tht, s with ny other theorem, one hs to mke sure tht ll the hypotheses hold before pplying it. The second is bit more subtle. One should not use it when it leds to circulr resoning, for exmple when the numertor or the denomintor of the limit L in question, which goes to 0 or s the cse might be, is differentible s needed, but to prove it one needs the limit L to exist in the first plce. Here is n exmple to illustrte this point. Consider the following two sttements: (I) The function e x is differentible with derivtive e x. (II) The limit equls. L = lim t 0 e t t 2

22 When presented with the limit L one is tempted to prove tht it is by using L Hôpitl s rule. Indeed, if we ccept tht it is pplicble here, then, since e t is by (I) differentible with derivtive e t, nd since the limit e t L = lim t 0 equls e 0 =, one thinks tht the problem is solved. But not so fst! This method ssumes (I) nd how does one prove it? Well, one hs to show the following: lim h 0 e x+h e x But since e x+h is e x e h, one hs to show tht lim h 0 h e h h = e x. =, which is the ssertion (II). So (I) nd (II) re equivlent nd one cnnot prove one using the other, unless one hs found different wy to prove one of them. So if one uses the L Hoˆpitl s rule to evlute L, one hs to show why e x is differentible without using L s tool, which cn be done. Of course one defines the exponentil function exp(x) s the inverse function of the logrithm nd the fct tht the derivtive of log x is /x implies, s we sw erlier, tht exp(x) is differentible with derivtive exp(x). But we then hve to show, by nother method, tht e x is the sme s exp(x). If one is not creful one will be drwn into delicte spider web. This is not to scre you into not using L Hôpitl s rule. Just mke sure before using it tht you cn stisfy the hypotheses nd tht there re no circulr rguments. Mke sure, in prticulr, tht the numertor nd the denomintor of the limit L cn be shown to be differentible without using L. Indeed, the most importnt thing one hs to lern in M is to think logiclly. Without further do, let us now present the rule of L Hôpitl. Proposition 8 (L Hôpitl s rule) Consider limit of the form L = lim x f(x) g(x), 22

23 where is either 0 or or or just ny (finite) non-zero rel number, nd f, g re differentible with g (x) 0 t ll rel numbers x with x sufficiently smll. Suppose both nd f nd g pproch 0, or both pproch, or both tend to, s x goes to. Then L exists if the limit quotient of the derivtives, nmely L = lim x f (x) g (x) exists. Moreover, when L exists, L equls L. Proof of L Hôpitl s Rule. We will prove this in the cse when =, with f(x), g(x) both pproching s x pproches. The other cses require only very slight modifictions nd will be left s exercises for the interested reder. By hypothesis, f(x) nd g(x) re defined nd differentible for lrge enough x, with g (x) 0 there. Suppose the limit L exists. We hve to show tht L lso exists, nd prove tht in fct L = L. The existence of L s (finite) rel number implies tht for every ε > 0, there is some b > 0 such tht for ll x > b, (A) f (x) g (x) L < ε. Since g(x) goes to s x, we my choose b lrge enough so tht g(x) g(b) for ll x > b. Applying Cuchy s MVT to (f, g) on [b, x], we get (A2) f (c) g (c) = f(x) f(b) g(x) g(b), for some c in (b, x). Combining with (A), we then get (A3) f(x) f(b) g(x) g(b) L < ε. In other words, (A4). f(x) f(b) lim x g(x) g(b) = L. 23

24 Now we re lmost there. To finish, note tht since f(x) s x, (A5) lim x f(x) f(x) f(b) =. Similrly, (A6) g(x) g(b) lim x g(x) =. The ssertion now follows by combining (A4), (A5) nd (A6). 24