Clculus Rigor, Concision, Clrity Jishn Hu, Jin-Shu Li, Wei-Ping Li, Min Yn Deprtment of Mthemtics The Hong Kong University of Science nd Technology
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Contents Rel Numbers nd Functions Rel Number System nd Inequlities Rel Number System Inequlities 3 Functions 6 3 Coordinte Plne nd Grphs of Functions 4 Summry 8 Limit nd Continuity 9 Limits of Sequences 9 Concept of Limits nd Properties 9 Rigorous Definition of Limit 6 Limits of Functions 37 Concept of Limit nd Properties 37 Rigorous Definition of Limit 44 3 Infinite Limit 5 3 Continuous Functions 54 4 Summry 6 3 Differentition 65 iii
iv CONTENTS 3 Derivtive 65 3 Computtion of Derivtive 7 33 Differentil 8 33 High Order Approimtion 83 34 Summry 87 4 Applictions of Differentition 89 4 Mimum nd Minimum 89 4 Men Vlue Theorem 95 43 L Hospitl s Rule 44 Incresing nd Decresing 4 45 Conveity 46 Summry 9 5 Integrtion 3 5 Riemnn Integrtion 3 5 Properties of Integrtion 3 53 Fundmentl Theorem of Clculus 33 54 Antiderivtive 4 55 Summry 5 6 Topics of Integrtion 53 6 Integrtion of Rtionl Functions 53 6 Numericl Integrtion 6 63 Improper Integrl 69 63 Integrl on Unbounded Intervl 69 63 Integrl of Unbounded Function 76 64 Summry 8 7 Applictions of Integrtion 85 7 Applictions in Geometry 85 7 Arc Length 85 7 Are of Surfce of Revolution 9 73 Volume of Solid of Revolution 96 74 Cvlieri s Principle 98 7 Polr Coordintes 4 7 Curves in Polr Coordintes 4 7 Length nd Are in Polr Coordintes 9
CONTENTS v 73 Physicl Applictions 4 73 Center of Mss 4 73 Work 9 74 Summry 8 Infinite Series 5 8 Series 5 8 Sum of Series 5 8 Comprison Test 8 83 Integrl Comprison Test 3 84 Absolute nd Conditionl Convergence 33 85 Rerrngement 36 8 Power Series 39 8 Tylor Series 39 8 Power Series 4 83 Opertion of Power Series 44 83 Fourier Series 47 83 Fourier Coefficients 47 83 Convergence of Fourier Series 5 84 Summry 5
Rel Numbers nd Functions REAL NUMBER SYSTEM AND INEQUALITIES In this section, we shll study the rel number system, which is the pltform for rel nlysis, nd how to solve inequlities Rel Number System We strt with the set of ll integers, Z It contins ll positive integers, negtive integers nd zero, ie, Z = {n n =, ±, ±, } In the set Z, we know tht there re few lgebric opertions llowed These re ddition, subtrction nd multipliction Division is generlly not llowed in Z, since it will generte frctions Besides the three lgebric opertions, nother property on Z is tht ny two numbers in Z cn be ordered In other words, if, b Z with b, then either < b or b < Becuse of the restriction of not llowing division, it is convenient to epnd the set Z to the set of ll rtionl numbers: Q = {r r = p/q, where p, q Z with q } Thus, rtionl number is quotient of two integers, with the denomintor not being zero Obviously, ny integer n is lso rtionl number, since n cn be epressed s n Hence, Z Q
REAL NUMBERS AND FUNCTIONS The order reltion between two rtionl numbers lso holds in Q However, there re mny interesting numbers not contined in Q For instnce, the length of the digonl of the squre with side length equls This number is not rtionl number Another interesting number is π, which is the rtio of the circumference of circle to the dimeter Therefore the number system Q is bit smll We would like to etend Q to bigger number system In fct, there is bigger number system R, clled the the field of rel numbers, contining numbers like nd π R hs mny similr properties s Q R hs lgebric opertions such s ddition, subtrction, multipliction nd division Numbers in R but not rtionl re clled irrtionl numbers The set R is lso ordered We do not discuss in detil the construction of rel numbers here, since it is beyond the scope of the book Thus, we hve three number systems, Z, Q nd R, stisfying Z Q R Besides the obvious difference tht there re irrtionl numbers which re in R, but not in Q, there is deeper difference between these two number systems In order to understnd the difference, we need to introduce the concept of the lest upper bound nd the gretest lower bound of subset of R Definition Given subset S of Q or R, we sy the set S is bounded bove, if there eists number u of Q (R respectively) such tht every number in S is no lrger thn u: s u, for ny s S Such u is clled n upper bound of S The smllest upper bound of S, if it eists, is clled the lest upper bound Similrly one cn define the concept of lower bound nd the gretest lower bound Here is n equivlent wy to understnd the lest upper bound We sy tht l is lest upper bound of S in Q (R respectively) if l is n upper bound of S nd for ny b < l nd b Q (R respectively), b is not n upper bound of S For instnce, the numbers,, 3 re ll upper bounds of the set (, ) in R The number is the lest upper bound of the set The fundmentl difference between Q nd R in the contet of nlysis is tht bounded set in R hs the lest upper bound in R, but bounded set in Q my not hve the lest upper bound in Q This cn be shown by nd S = {q Q q } Q, T = {r R r } R In fct, we hve the following generl theorem, which will not be proved here Theorem Let S be subset of R bounded bove Then S hs the lest upper bound We sy tht the rel number system hs the lest upper bound property will use this theorem without proof In this course, we
REAL NUMBER SYSTEM AND INEQUALITIES 3 From the previous emple, we see tht the number system Q does not hve this property We use the coordinte line (or rel line) to represent the rel number system The coordinte line is line with distinguished point O, clled the origin, nd n right rrow on the right end indicting the positive direction of the line A point on the rel line is represented by rel number Points on the left (right resp) side of the origin represent negtive (positive resp) rel numbers The origin represents the zero These re shown in Fig O Fig The rel line Inequlities Given two rel numbers nd b with < b, the set is clled n open intervl The set (, b) = { R < < b} [, b] = { R b} is clled closed intervl There re other kinds of intervls: (, b] = { R < b}, [, b) = { R < b}, (, + ) = { R < }, [, + ) = { R }, (, b) = { R < b}, (, b] = { R b} The whole rel line (, + ) = R These intervls re specil cses of subsets of R There re opertions on subsets: union, intersection nd complement Given two subsets S nd T of R, the union S T is defined by the intersection S T is defined by nd the complement S c is defined by S T = { R S or T }, S T = { R S nd T }, S c = { R / S} Here is the intuitive wy to eplin the words open nd closed The intervl (, b) is open since ll the inequlities re ll strict inequlities The intervl [, b] is closed since ll the inequlities re either or
4 REAL NUMBERS AND FUNCTIONS The precise definition of n open set in R cn be stted s follows A set S R is open if for ny point p S, there eists (, b) S contining p A set S is closed if the complement of S is open There is n subset of R, clled the empty set, which does not contin ny elements of R For emple, the intersection (, ) (, ) = According to the definition, R is open nd is closed since is the complement of R The empty set is lso open ccording to the definition In fct, R nd re the only two sets tht re both closed nd open Let us recll some properties of inequlities If < b, then + c < b + c for ny c R; If < b nd c < d, then + c < b + d; 3 If < b nd c >, then c < bc Emple The inequlity + 3 < 3 + 8 cn be solved by moving terms round to hve n equivlent inequlity 5 < 5 Multiply both sides by /5 nd switch the direction of the inequlity, we get the solution > Emple To solve the inequlity 3 + >, we fctorize the polynomil first: 3 + = ( )( ) The inequlity is equivlent to the fct tht nd hve the sme signs, ie, > nd > ; or < nd < We obtin > or < A more intuitive wy to solve the inequlity is s follows One should observe tht the monomils nd chnge sign from negtive to positive t nd, respectively Since the roots nd of the polynomil 3 + = ( )( ) divides the rel line into three open intervls (, ), (, ), nd (, + ), we see tht this polynomil chnges sign when moves from one intervl to n djcent one, s demonstrted in Fig In other words, the polynomil ( )( ) tkes the positive sign in the intervl (, ), subsequently negtive sign in (, ) nd positive sign in (, + ) Therefore the solution for the originl inequlity is tht lies in the intervl (, ) or the intervl (, + ) Equivlently, one gets <, or >
REAL NUMBER SYSTEM AND INEQUALITIES 5 + + + + + + + + + + + + + + + + + + + + ( )( ) Fig Signs of the polynomils,, nd ( )( ) Emple 3 The inequlity < < < cn be solved by using the sme ide s bove to get Emple 4 To solve the inequlity 3 3 >, we first chnge it to n equivlent inequlity ( )( ) > The roots of the polynomil re, nd They divide the rel line into intervls (, ), (, ), (, ) nd (, + ) The polynomil ( )( ) tkes negtive sign when is very negtive Hence ( )( ) is negtive in (, ), positive in (, ), negtive in (, ), nd positive in (, + ), s shown in Fig 3 The solution for the inequlity is tht is in (, ) or (, + ) Equivlently < < nd > + + + + + + Fig 3 Signs of the polynomil ( )( ) Eercises Show is n irrtionl number Show 3 is n irrtionl number 3 Solve the following inequlities:
6 REAL NUMBERS AND FUNCTIONS 3 4 < ; 6 ( )( ) ( 3) 3 ( 4) 4 < ; < 3; 3 > ; 4 3 + 4 > 4 ; 5 ( )( )( 3)( 4) > ; 7 8 + + > ; ( ) > + FUNCTIONS Given subset D of R, function is rule tht ssigns rel number f() to ny D For emple, the function f() = is the rule tht ssigns the squre of for ny in D = R The subset D is clled the domin of the function The letter is clled the independent vrible Thus, the domin of function is the set of ll llowble vlues of the independent vrible of the function We often write y = f() for the function f() The letter y is clled the dependent vrible The set of possible vlues of dependent vrible is clled the codomin The rnge of function is the set of ll ctul vlues the function my return corresponding to the domin In other words, the rnge is the set f(d) Obviously, it is lwys subset of the codomin D f : D C f(d) C Fig 4 An illustrtion of function f : D C Often we use the nottion f : D C, where D is the domin, nd C is the codomin This nottion implicitly suggests tht f(d) C In mny cses, function hs n eplicit epression such s f() = The function f() = cn be defined on the whole R or ny subset of R However there re mny functions whose domins re smller thn the whole R For emple, the lrgest possible domin of f() = is the intervl [, + ) Sometimes, the domin of function my not be given eplicitly If function hs n eplicit epression, we my find the lrgest possible domin of f() in mny cses For emple, the lrgest possible domin of the function f() = is the rel line with the origin deleted Emple The function f() = 3 + is meningful when 3 +
FUNCTIONS 7 From Emple, we get, or Thus the domin is the union of two intervls (, ] [, + ) Here re emples of functions we will encounter frequently in this book (i) Polynomil functions f() = n + n + + n + n The lrgest possible domin for polynomil function is the rel line (ii) Rtionl functions F () = f() g(), where both f() nd g() re polynomils The lrgest possible domin for F () is the rel line with roots of g() deleted (iii) Algebric functions re functions obtined from polynomils vi ddition, multipliction, division nd tking roots For instnce, + + is n lgebric function (iv) Trnsendentl functions stnd for those which re not lgebric functions For emple, e, ln, sin, cos, tn nd cot re ll trnsendentl functions (v) There re lso functions epressed in pieces such s f() = {, if <,, if >,, if <, nd g() =, if, 3, if > Here re mens to generte more vrieties of functions Given function f with domin A nd function g with the domin B, there is function h defined by h() = f() + g() The domin of h is the intersection A B We usully denote the function h by f + g nd write Similrly, we cn define new function f g by with the domin A B; new function f/g by (f + g)() = f() + g() (f g)() = f() g() (f/g)() = f()/g()
8 REAL NUMBERS AND FUNCTIONS with the domin nd new function cf by A { B g() }; (cf)() = cf() with the domin A for constnt c If the rnge of f is contined in the domin B of g, we cn define the composition h of g nd f by h() = g(f()) Usully we use g f to denote the composition h Emple Let us look t some more complicted functions Suppose f() = e, g() = sin, nd h() = The composition function f g is The composition function g f is f g() = f(g()) = e sin g f() = sin e We see tht f g g f The composition function h (f g) is The composition function (f g) h is h (f g)() = h(f g()) = (e sin ) = e sin (f g) h() = f g(h()) = e sin Inverse Functions A function f : D C, with the domin D nd the codomin C, is clled one-to-one, or injective, if f( ) f( ) whenever The term one-to-one is in contrst to mny-to-one A function f : D C is clled from D onto C, or surjective, if for every y C, there eists t lest one D such tht f() = y In fct, f is surjective iff (if nd only if) C is the rnge of f, tht is, C = f(d) If f : D C is injective s well s surjective, then it is clled bijective Let f : D R be bijective function We cn define new function g whose domin is R s follows For ech r R, there eists unique d in D such tht f(d) = r We define new function g so tht g(r) = d The function g hs the property tht nd f g() =, for ny R When we write f : D R, we implicitly men tht the codomin R is the rnge f(d) (R for rnge ) If we do not intend to emphsize this distinction, we will use the letter C in generl One cn lwys write the codomin of ny function to be R = (, + ), since it contins ll rel numbers When the domin of function is not shown eplicitly, it is usully the set where the function is defined For emple, for f() =, the domin is
FUNCTIONS 9 g f() =, for ny D The function g is clled the inverse function of f, sometimes denoted by f Emple 3 Consider f() = + b, with d bc c + d Let us first ssume tht c It is esy to see tht the domin of the function f is R\{ d/c} However, it is not so obvious wht the rnge of f is The inverse function of f cn be found s follows First, by writing y = + b nd solving for c + d to hve = dy b cy We switch the symbols of the dependent nd independent vribles to hve the inverse function g() = d b c, whose domin is R\{/c} This suggests tht the function f() = + b is bijective from R\{ d/c} to R\{/c} In fct, c + d we cn show tht it is both injective nd surjective Injective If, we hve which implies Since c + d nd c + d, we hve Tht is, the function f is injective d( ) bc( ), ( + b)(c + d) ( + b)(c + d) + b c + d + b c + d Surjective For ny y R\{/c}, tke = dy b + b Then for such, we hve y = cy c + d This mens tht the function f is surjective Now let us emine the compositions of f nd g Elementry clcultions give ( ) d b c + b ( )(d b) + b(c ) f g() = ( ) = c + d ( c)(d b) + d(c ) =, d b c for ny R\{/c}; If d = bc, then the function is constnt
REAL NUMBERS AND FUNCTIONS nd ( d g f() = ( c +b c+d +b c+d ) b ) d( + b) b(c + d) = c( + b) (c + d) =, for ny R\{ d/c} If c =, then, d, since d bc In this cse, it is esy to show tht f is liner function nd bijective from R to R Emple 4 The domin of the function f() = e is D = R nd the rnge is R = (, + ), while the codomin C cn be either [, + ) or R It is bijective from R to (, + ) The function g() = ln : (, + ) R is the inverse of f, since the following two identities hold: f g() = e ln =, for ny R = (, + ) nd g f() = ln(e ) =, for ny D = (, + ) The function f() = : D = R [, + ) hs no inverse function since it is not one-to-one For instnce, f( ) = = f() However the function h() = : [, + ) [, + ) hs n inverse h () = The trigonometric functions sin, cos, tn nd cot re periodic functions If we only consider the sine function sin over the intervl ( π/, π/), then it is one-to-one nd hence hs n inverse, denoted by rcsin defined over (, ) Similrly, the cosine function cos defined over (, π) hs n inverse function rccos over (, ) The function tn defined over ( π/, π/) hs n inverse rctn defined over the whole rel line The function cot defined over (, π) hs n inverse rccot defined over the whole rel line We sy function is incresing if is decresing if f( ) > f( ) when > ; f( ) < f( ) when > Both re clled monotonic functions It is esy to see tht ny monotonic function is one-to-one nd hence hs inverse function from the domins to its rnge Emple 5 The function f() = 3 is incresing The function g() = sin is incresing in the intervl ( π/, π/) The function h() = cos is decresing over the intervl (, π) Eercises Determine the domins nd rnges of the following functions:
3 COORDINATE PLANE AND GRAPHS OF FUNCTIONS f() = ; f() = 3 ; 3 f() = 3 ; 4 f() = + ; 5 f() = ln( ); 6 f() = ln( + ) + ln( ); 7 f() = sin ; 8 f() = ln(sin ) Show tht ech of the functions is incresing in the specified intervl: f() = ( < + ); f() = sin ( π π ) ; 3 f() = tn ( π < < π ) ; 4 f() = + sin ( < < + ) 3 Show tht ech of the functions is decresing in the specified intervl: f() = ( < ); f() = cos ( π) ; 3 f() = cot ( < < π) ; 4 f() = ( )( ) ( < 3/) 4 Determine the region where the function is monotonic: f() = 3 ; f() = + b + c; 3 f() = + ; 4 f() = 4 5 Find the inverse function: f() = + ; f() = ( < ); 3 f() = ( < + ); 4 f() = + ( ); 5 f() = ( ); 6 f() = ( ) 3 COORDINATE PLANE AND GRAPHS OF FUNCTIONS Drw horizontl line with right rrow on the right end nd verticl line with n upwrd rrow on the top end The two perpendiculr lines intersect t one point nmed the origin The two lines re rel lines The horizontl line is clled the -is The verticl line is clled the y-is Ech point P on the plne corresponds to pir (, b) s follows Project the point verticlly towrds
REAL NUMBERS AND FUNCTIONS y b P = (, b) O Fig 5 The coordinte plne the -is to get point on the -is Similrly project the point horizontlly towrds the y-is to get point b on the y-is The points nd b re clled -coordinte nd y-coordinte respectively This pprtus is clled the coordinte plne y y P = (, y ) y P = (, y ) y y R = (, y ) Fig 6 The distnce from P = (, y ) to P = (, y ) Given two points P = (, y ) nd P = (, y ) Without loss of generlity, ssume < nd y < y Let R = (, y ) We get right tringle P P R The edge P R hs length equl to The edge RP hs length equl to y y The distnce from P to P equls the length of the edge P P Therefore we hve the distnce formul distnce from P to P = (y y ) + ( ) Given function y = f() For ech, we get point (, f()) on the coordinte plne When moves in the domin D of f, the point (, f()) moves on the plne The trce of (, f()) for D is clled the grph of f Lines nd Liner Functions Emple 3 Consider the liner function f() = + b We know tht the grph of f is stright line, shown in Fig 7 When = = b, the grph of the function f is the -is, line
3 COORDINATE PLANE AND GRAPHS OF FUNCTIONS 3 y b O y = + b Fig 7 The grph of the stright line y = + b y L P L O θ P Fig 8 A line is determined by the coordintes of two points In the generl cse, s in Fig 8, given ny two distinct points P = (, y ) nd P = (, y ) on the grph of f, we hve y = + b nd y = + b y y = ( + b) ( + b) = ( ), = y y Therefore is determined by (, y ) nd (, y ) Subsequently b is lso determined by these two points: b = y The coefficient in f() = + b is clled the slope of the line nd b is the y-intercept of the line since the intersection of the L with the y-is is (, b) The slope of line is very importnt concept In the emple bove, s shown in Fig 8, let θ be the ngle between the line L nd the line L It is esy to see tht tn θ = y y = Given two non-verticl lines L nd L Let i be the slope of L i for i =, Suppose L nd L re perpendiculr, s in Fig 9 Let θ be the ngle between the -is nd L nd θ be the ngle between the -is nd L Then θ θ = ±π/, cos(θ θ ) =, cos θ cos θ + sin θ sin θ =
4 REAL NUMBERS AND FUNCTIONS y L θ θ L Fig 9 Two perpendiculr stright lines Therefore we get = tn θ tn θ = y L L θ θ Fig Two prllel stright lines Given two non-verticl lines L nd L If they re prllel, s in Fig, then the ngel between L nd the -is is equl to the ngle between L nd the -is Therefore the slopes of two lines re the sme: = The verticl line cn be describe by the eqution = k, s shown in Fig y = k Fig Verticle line = k
3 COORDINATE PLANE AND GRAPHS OF FUNCTIONS 5 The coordinte plne provides pltform to describe geometric objects such s lines nd curves by equtions F (, y) = of two vribles nd y For emple, the grph of function f() corresponds to the eqution F (, y) = y f() =, s in Fig f() (, f()) Fig The grph of function F (, y) = y f() = Circles, Ellipses, Hyperbols nd Prbols The circle of rdius r centered t the point (, b) is represented by the eqution ( ) + (y b) = r from the distnce formul, shown in Fig 3 y r (, b) Fig 3 The circle ( ) + (y b) = r y b (, y ) Fig 4 The ellipse ( ) (y y) + = b The eqution ( ) + (y y ) b =
6 REAL NUMBERS AND FUNCTIONS describes curve clled n ellipse, shown in Fig 4 The point (, y ) is clled the center of the ellipse b ( ) (y y) = b b ( ) (y y) + = b Fig 5 Hyperbols Both re centered t (, y ) The eqution ( ) (y y ) b = describes curve clled hyperbol The eqution ( ) + (y y ) b = is nother hyperbol Both re shown in Fig 5 The grph of the function f() = is clled prbol The curve psses the point (, ) tht is clled verte of the prbol It opens upwrds or downwrds depending upon the sign of The eqution y = provides nother kind of prbol opening rightwrds if > nd leftwrds if < These re demonstrted in Fig 6 y y y y = y = = y > < > Fig 6 Prbols Let C be the curve corresponding to the eqution F (, y) = The curve F (, y b) = is obtined by moving the curve C horizontlly by nd verticlly by b, equivlently, moving the coordinte plne horizontlly by nd verticlly by b, shown in Fig 7 It is clled trnsltion of F (, y) = In Fig 8, the curve F (, y) = is obtined by reflecting the curve C with respect to y-is The curve F (, y) = is obtined by reflecting the curve C with respect to -is
3 COORDINATE PLANE AND GRAPHS OF FUNCTIONS 7 y y O O (, b) Fig 7 Trnsltion of curve: from F (, y) = to F (, y b) = y F (, y) = O F (, y) = y O F (, y) = F (, y) = Fig 8 Reflections of curve F (, y) = with repect to both ises Emple 3 To sketch the grph of f() = + 4 + 4, we rewrite it s or y = ( + ) +, y = ( ( )) The grph of the function f is the prbol y = with the verte moved to (, ), s shown in Fig 9 y y = + 4 + 4 (, ) Fig 9 Trnsltion of y = Eercises 3 Grph the liner function f() = for =, /,, nd
8 REAL NUMBERS AND FUNCTIONS 3 Grph the liner function f() = + b for b =,, nd 33 Grph the liner functions: () f() = + ; (b) f() = ; (c) f() = ; (d) f() = 34 Grph the following equtions: + y 4 + y + 3 = ; 4 + 4y + 8 4y + 3 = ; 3 + y 4 + y + = ; 4 3 + y + 3 + y = ; 5 + y = ; 6 + + y = ; 7 4 + y + 5 = ; 8 4y + 4 4y + 3 = ; 9 y 4 + y + = ; 3 y + 3 + y = 4 SUMMARY Definitions A function f : D C, with the domin D nd the codomin C, is clled one-to-one, or injective, if f( ) f( ) whenever A function f : D C is clled from D onto C, or surjective, if for every y C, there eists t lest one D such tht f() = y A function f : D C is clled bijective, if it is injective s well s surjective A function f is incresing if f( ) > f( ) when > ; is decresing if f( ) < f( ) when > Given subset S R, the set S is bounded bove, if there eists number u R such tht every number in S is no lrger thn u: s u, for ny s S Such u is clled n upper bound of S The smllest upper bound of S, if it eists, is clled the lest upper bound Theorems Lest Upper Bound Property Let S be subset of R bounded bove Then S hs the lest upper bound
Limit nd Continuity LIMITS OF SEQUENCES Limit is one of the most bsic concepts in clculus We strt with limit of sequences nd lter, we will study limit of functions Concept of Limits nd Properties We introduce the concept of sequences nd their limits, nd discuss properties of limits We will concentrte on the ides, leving the rigorous tretment to the lter subsection A sequence { n } is,,, n, We cll n the n-th term of the sequence In this course, we will only consider sequence of numbers, which mens tht n re rel numbers Here re some emples n = n:,, 3,, n, ; b n = :,,,,, ; c n = n :,,, n, ; d n = ( ) n :,,,, ( ) n, ; e n = sin n: sin, sin, sin 3,, sin n, Note tht the inde n in sequence does not hve to strt from For emple, the sequence {d n } ctully strts t n = (or ny even integer) 9
LIMIT AND CONTINUITY Let us compre these sequences When n gets lrger, n gets lrger, b n is constnt, c n gets closer to zero, d n nd e n does not pproch nything The sequences {b n } nd {c n } shring the property of pproching finite number L (L = for b n nd L = for c n ) when n goes to infinity We sy the sequence converges to the limit L In contrst, we sy the sequences { n }, {d n } nd {e n } diverge since they do not pproch ny finite number when n goes to infinity Definition (Non-rigorous) A sequence { n } converges to finite number L if n pproches L when n goes to infinity We write lim n n = L, nd cll the sequence { n } convergent We cll sequence divergent if it does not pproch ny finite number when n goes to infinity Since the limit describes the behvior when n gets very lrge, the modifiction of finitely mny terms in sequence does not chnge the convergence nd the limit vlue Intuitively, we know tht if is close to 3 nd b is close to 5, then + b is close to 3 + 5 = 8 The intuition leds to the following properties Proposition (Arithmetic Rule) Suppose { n } nd {b n } converge Then { n +b n }, {c n }, { n b n } converge (c denotes constnt) nd lim ( n + b n ) = lim n + lim b n, n n n Moreover, if lim n b n, then { n b n lim c n = c lim n, n n } lso converges, nd lim n n n lim = n b n lim b n n Proposition 3 If { n } converges, then { n } lso converges nd the other hnd, if lim n n =, then lim n n = lim nb n = lim n lim b n n n n lim n = n lim n n On Emple Bsed on the limits lim c = c, lim = nd the rithmetic rule, the limit of { n n n n } + n the sequence n my be computed s follows n + n + n + ( lim + ) lim n n n + = lim n n n + lim n n n + = ( n n + n ) = n = lim n Emple We hve lim = nd = n n n ( ) n lim = n n lim n n + lim n n lim n ( ) n n By Proposition 3, we get n
LIMITS OF SEQUENCES The following property is very useful for deriving more sophisticted limits Proposition 4 (Sndwich Rule) If n b n c n nd lim n = lim c n = L, then lim b n = n n n L The rule reflects the intuition tht if nd c re close to 5, then nything between nd c should lso be close to 5 Emple 3 By sin n nd the sndwich rule, we get lim n n = n sin n n n, lim n n =, Emple 4 To compute the limit lim n ( n + n), we note tht < n + n = ( n + n)( n + + n) = < n + + n n + + n n By the sndwich rule nd lim n n = lim n Emple 5 Let > We show tht In the cse, we let α n = n Then nd we get = ( + α n ) n = + nα n + n =, we get lim n ( n + n) = lim n = n n(n ) αn + + αn n + nα n, α n n By the sndwich rule nd lim = ( ) lim =, we get lim n n n n α n = nd n lim n = lim α n + = n n If <, we hve By the rithmetic rule, lim n = lim n n n = lim n n =
LIMIT AND CONTINUITY ( Emple 6 To compute the limit lim ) sin n, we note tht n n 3 ( By lim ) = n n 3 + ( n 3 ) sin n n n 3 3 lim n Emple 7 To compute the limit Then ( n = nd the sndwich rule, we get lim ) sin n = 3 n n 3 n n + n + lim, we note tht < n n n n + < n n < n 6 n n + nd the sndwich rule tell us lim n n = Emple 8 We show the limit lim For n > 5, we hve n n n! = for the specil cse = 5 < 5n n! = 5 5 5 5 5 3 4 5 5 6 5 7 5 n 5 5 5 5 5 3 4 5 5 n = 55 4n 5 5 By lim n 4n = 55 4 lim 5 n = nd the sndwich rule, we get lim n n n n! = Emple 9 Let < We show the limit lim n n = For < <, write = Then b > nd + b < n = ( + b) n = n(n ) + nb + b + + b n nb < 6 for n > By lim = nd the sndwich rule, we get lim n nb n n = If < <, we hve n n n By lim n n = nd the sndwich rule, we lso get lim n n = A sequence { n } is bounded bove if there is number B (clled n upper bound), such tht n B for ll n The sequence is bounded below if there is number B (clled lower bound), such tht n B for ll n A sequence is bounded if it hs both upper nd lower bounds Proposition 5 Any convergent sequence is bounded The property reflects the intuition tht if is close to π, then is between 3 nd 4
LIMITS OF SEQUENCES 3 { n + ( ) n } Emple The sequences {n}, diverge becuse they re not bounded bove n + On the other hnd, the sequence,,,, is bounded nd diverges Therefore the converse of Proposition 5 does not hold The following property reflects the intuition tht lrger number should be closer to lrger limit Proposition 6 (Order Rule) Suppose { n } nd {b n } converge If n b n, then lim n n lim n b n If lim n n > lim n b n, then n > b n for sufficiently lrge n On the other hnd, the condition n > b n does not necessrily imply lim n n > lim n b n This cn be demonstrted by n = n nd b n = n n + n Emple By lim n n n + fct, the inequlities hold for ll n n + n By lim n holds for n > n n =, we know lim n =, we know < n + n n < 3 for sufficiently lrge n In n + n + < for sufficiently lrge n In fct, the inequlity n Monotonic Sequences A sequence { n } is incresing if It is decresing if Both re monotonic { } sequences { The sequences, n n 3 n n+ 3 n n+ }, { n } re decresing The sequences Proposition 7 Any bounded monotonic sequence converges { }, {n} re incresing n An incresing sequence { n } is bounded if it hs upper bound, becuse the first term of the sequence is lredy the lower bound Similr remrk cn be mde for decresing sequence Emple The sequence, +, + +, is inductively given by =, n+ = + n
4 LIMIT AND CONTINUITY We clim tht the sequence { n } is incresing nd bounded bove by We hve = < nd = + > Now ssume k < nd k+ > k Then k+ = + k < + =, k+ = + k+ > + k = k+ The clim is proved by induction By Proposition 7, the sequence converges Let L be the limit Then by tking the limits on both sides of the equlity n+ = + n nd pplying the rithmetic rule, we get L = + L Therefore L = or Since n >, by the order rule, we must hve L Therefore we conclude tht lim n n = Emple 3 We will show tht the limit lim n ( + ) n n converges We compre two consecutive terms by the binomil epnsion, ( + ) n = + n n(n ) n(n ) 3 + + + n n! n n! n n = +! + ( ) + + ( ) (! n n! n n ( + ) n+ = + n +! + ( ) + (! n + n! n + + (n + )! ( n + ) ( n + ) ) ) ( ( n n + ( n n n + ) ) ), ( n ) n + A close emintion shows tht the sequence is incresing It remins to show tht the sequence hs upper bounded By the epnsion bove, we hve ( + ) n < + n! +! + + n! < + + + 3 + + = + + ( ) + = + + n < 3 ( 3 (n )n ) + + By Proposition 7, the sequence converges The limit is denoted by e ( + n) n = e lim n ( n ) n Emple 4 We give nother rgument tht lim n n = for < <
LIMITS OF SEQUENCES 5 Since < <, the sequence { n } is decresing nd stisfies < n < Therefore the sequence converges to limit L On the other hnd, the sequence { n } is obtined by deleting the first term from { n } Therefore the sequence { n } should lso converge to the sme limit L If L, then we my pply the rithmetic rule to get = lim n n n = lim n n lim = L n n L = This contrdicts to the ssumption tht < Therefore the limit L hs to be Subsequences By choosing infinitely mny terms from sequence { n }, we get subsequence The newly formed sequence usully is written s { nk }, ie, The indices stisfy n, n,, nk, n < n < < n k <, nd n k k, for every k { For emple, the following re subsequences of n = }, corresponding to the choices n k = k, n n k = k, n k = k, n k = k!: k = k :, 4, 6,, k, ; k = k :, 3, 5,, k, ; k = k :, 4, 8,, k, ; k! = k! :,, 6,, k!, Proposition 8 If sequence converges to L, then every subsequence converges to L The property reflects the intuition tht if something is close to L, then ny pert of it is lso close to L n + n Emple 5 By lim n n =, we lso know n + lim n (n ) + (n ) (n ) (n ) + = lim n 4 n + n n 4 3n + 3 =,
6 LIMIT AND CONTINUITY nd lim n (n!) + n! (n!) n! + = Emple 6 Consider the sequence { n = ( ) n } The subsequence { k = } converges to, nd the subsequence { k = } converges to Since the two subsequences hve different limits, Proposition 8 tells us tht the sequence {( ) n } diverges The concept of subsequences llows us to stte the prtil converse to Proposition 5 Theorem 9 (Bolzno-Weierstrss) theorem, Bolzno-Weierstrss Every bounded sequence hs convergent subsequence As typicl emple, the sequence {( ) n } is bounded Although the whole sequence diverges, the sequence hs two subsequences converging to nd respectively Emple 7 Let us list ll the rtionl numbers in (, ] s sequence =, =, 3 =, 3, 3, 3 3,, n, n,, n n, The number π = 383988684 is the limit of the sequence 3, 3, 38, 383, 383, 3839,, which is clerly subsequence of { n } It is esy to see tht ny number in [, ] is the limit of convergent subsequence of { n } Rigorous Definition of Limit The intuitive definition of limit is mbiguous becuse the mening of the phrses pproches L nd goes to infinity is not precise To rrive t more precise definition, we study the sttement lim n n = in more detil When we sy tht n = pproches L = s n goes to infinity, we n men n infinite collection of fcts n > = n L <, n > = n L <, n > = n L <, n > = n L <, In everydy life, we mesure the lrgeness of quntity by compring with some specific scle For emple, city is considered s big if it hs millions of people (n > ), nd the country is
LIMITS OF SEQUENCES 7 considered s big if it hs hundreds of millions of people (n > ) On the other hnd, electric wire is considered thin if it is less thn millimeter in dimeter (d < ), nd hir my be considered s thin if it is less thn 5 millimeter in dimeter (d < 5) Thus in ordinry lnguge, we sy when n is in the hundreds, then n L is in the hundredth, nd when n is in the millions, then n L is in the millionth, etc Of course for different limit, the reltion between the bigness of n nd the smllness of n L my be different For emple, for n = n nd L =, we hve n! n > = n L < 3, n > = n L < 5, But the key observtion here is tht limit is n infinite collection of sttements of the form if n is lrger thn certin lrge number N, then n L is smller thn certin smll number ɛ In prctice, we cnnot verify ll such sttements one by one Even if we hve verified the truth of the first one million sttements, there is no gurntee tht the one million nd the first sttement is true To mthemticlly estblish the truth of ll the sttements, we hve to estblish the sttements for ll N nd ɛ t once Which one sttement do we need to estblish? We note tht n > N is the cuse of the effect n L < ɛ, nd the precise reltion between the cuse nd the effect vries from limit to limit For some limit, n in thousnds lredy gurntees tht n L < For some other limit, n hs to be in the billions in order to mke sure tht n L < But the key observtion here is tht no mtter how smll ɛ is, the trget n L < ɛ cn lwys be gurnteed for sufficiently lrge n So lthough n L < my not be stisfied for n in the million rnge, it will probbly be stisfied for n in the billions rnge (nd if this fils gin, perhps in billions nd billions rnge) In fct, for ny given rnge ɛ >, we cn lwys find n integer N such tht n is close to the limit L within the rnge ɛ whenever n N, ie, n L < ɛ, when n N In the cse of n =, we cn tke N to be ny integer greter thn /ɛ One of such integer could n be [/ɛ] +, where [] represents the integer prt of the number The concept of limits is chrcterized by the property tht no mtter how smll (or close) the trget is set, we cn lwys chieve the trget by going fr enough in the sequence to chieve the trget Definition (Rigorous) A sequence { n } converges to finite number L if for ny ɛ >, there is n integer N, such tht n > N implies n L < ɛ { } [ ] Emple 8 Consider the sequence For ny ɛ >, choose N = + When n > N, n ɛ we then hve n = n < N = ɛ Any number cn be epressed s = [] + (), where [] is the integer prt of nd () is the frctionl prt
8 LIMIT AND CONTINUITY This proves lim n n = rigorously Emple 9 Consider the sequence { n }, with n = n n for ech n For ny ɛ >, we cn + choose N to be ny integer greter thn Whenever n N, ɛ n n + = n + N + /ɛ + = ɛ + ɛ < ɛ Thus the sequence converges to How nd why did we choose? The ide is tht we need ɛ n n + < ɛ This inequlity is equivlent to the inequlity n > ɛ Therefore, ny integer greter thn cn be choose to be N Above, we chose N ɛ ɛ We 4 cn lso choose N ɛ However, the description bove hs n obvious flw tht requires to solve the inequlity n n + = n + < ɛ Generlly, this is not fesible, since some inequlities cn be too difficult to solve One wy to overcome this is to loose the inequlity until it cn be solved esily The key to do this is to keep trck of every condition whenever you loose the inequlity In fct, we cn solve the bove inequlity s follows In the sequence of inequlities: n n + = n < 3 ɛ, + n N the inequlity holds for ll nturl integers n; the inequlity holds under the condition n N; the lst inequlity 3 holds under the condition tht N Thus, the whole sequentil inequlities ɛ hold when n N, N is n integer greter thn ɛ Hence, n n + < ɛ
LIMITS OF SEQUENCES 9 when Therefore, by the definition, we hve n [ ] + ɛ n lim n n + = The bove loose-nd-trck pproch is very useful in proving eistence of limits Emple We demonstrte ( similr rgument once more by using the loose-nd-trck pproch for the limit lim n + n) In fct, for the sequence of inequlities, n n + n = ( ) ( n + + n ) n + n ( n + + n ) = n + + n < n N the inequlity holds for ll nturl integers n; the inequlity holds under the condition n N; the lst inequlity 3 holds under the condition tht N Thus, the whole sequentil inequlities ɛ hold when 3 ɛ, n N, N is n integer greter thn ɛ Hence, n + n < ɛ when Therefore, by the definition, we hve n [ ] + ɛ [ lim n + n] = n In the emples bove, we sw tht we often need to loosen up the difference between the sequence nd the epected limit Such estimtion llows us to derive simple choice of N tht stisfies the definition Of course the wy of loosen up is not unique For emple, the estimtion n + n n n + = 3n n n + < 3n n n + < 3n n n = 6 n
3 LIMIT AND CONTINUITY will led to the choice N = 6 ɛ nd the estimtion n + (n + ) n n = < = n + + n n + n n [ ] will led to the choice N = + So the choice of N is not unique The only importnt thing ɛ here is tht some N cn be found to stisfy the requirement of the definition, nd N depends only on ɛ (nd especilly independent of n) The rigorous definition of limit llows us to rigorously prove some properties of limit Emple Assume lim n = n = We prove tht lim n n The ssumption tells us tht for ny ɛ >, there is n integer N, such tht n > N implies n < ɛ Then n > N = n = n n + n < ɛ This completes the rigorous proof By the similr method, we cn rigorously prove tht lim n = L > implies lim k n = k L n n for ny nturl number k In fct, we lso hve lim n p n = L p for ny number p But the proof would be more complicted Emple [Prove of lim ( n + b n ) = lim n + lim b n] n n n Assume lim n = L nd lim b n = K For ny ɛ >, we still hve ɛ n n definition of the limits of { n } nd {b n } to ɛ, we find N nd N, such tht > Applying the n > N = n L < ɛ, n > N = b n K < ɛ Then we get n > N = m{n, N } = ( n + b n ) (L + K) n L + b n K < ɛ + ɛ = ɛ This completes the proof tht lim n ( n + b n ) = L + K If pply the definition to ɛ insted of ɛ, then we will get n > N implies ( n + b n ) (L + K) < ɛ in the end It turns out tht replcing < ɛ by < ɛ (or by < ɛ, or by < ɛ, or by ɛ, etc) in (the finl prt of) the definition of limit gives new definition tht is equivlent to the old one Emple 3 [Prove of Proposition 4] Assume n b n c n nd lim n = lim c n = L n n For ny ɛ >, there re integers N nd N, such tht n > N = n L < ɛ, n > N = c n K < ɛ Then n > N = m{n, N } = L ɛ < n b n c n L < L + ɛ = b n L < ɛ
LIMITS OF SEQUENCES 3 This proves tht lim n b n = L Emple 4 [Proof of Proposition 5] Assume lim n n = L By the definition of limit, for ɛ =, there is n integer N, such tht n > N implies n L < By choosing lrger N if necessry, we my further ssume N to be nturl number Then we hve N+ < L +, N+ < L +, N+3 < L +, Tke B = m{,, N, L + } Then we get n B for ll n Therefore the sequence is bounded by B Note tht if in the definition of limit, N is not n integer (or nturl number), then the new definition of limit is still equivlent to the old one In the current emple, it is more convenient to ssume N to be nturl number here becuse we need to use N s the inde of term in sequence Emple 5 [Prove of lim ( nb n ) = ( lim n) ( lim b n)] n n n By Proposition 5 just proved, { n } nd {b n } re bounded Assume n < M nd b n < M Assume lim n = L, lim b n = L n n For ny ɛ >, there is n integer N such tht nd there is n integer N such tht n > N = n L < n N = b n L < Choose N = m{n, N } Then, if n N, we hve n b n L L ɛ M, ɛ L + = n b n L b n + L b n L L = ( n L )b n + L (b n L ) b n n L + L b n L ɛ ɛ < M + L M L + < ɛ Emple 6 [Proof of Proposition 6] Assume n b n nd lim n = L nd lim b n = K n n For ny ɛ >, there re N nd N, such tht n > N = n L < ɛ, n > N = b n K < ɛ Now pick nturl number n lrger thn both N nd N Then we hve L + ɛ > n b n > K ɛ
3 LIMIT AND CONTINUITY Therefore we proved tht L + ɛ > K ɛ for ny ɛ > It is esy to see tht the property is the sme s L K Conversely, we ssume lim n = L, lim b n = K nd L > K For ɛ = L K >, there re N n n nd N, such tht n > N = n L < ɛ, n > N = b n K < ɛ Then for n > N = m{n, N }, we hve Therefore n > b n s long s n > N n b n > (L ɛ) (K + ɛ) = L K ɛ = Emple 7 [Proof of Proposition 8] By the definition, for ny ɛ >, there is n integer N so tht n L < ɛ whenever n N Thus, for ny subsequence { nk }, if k N, since n k k N, we hve nk L < ɛ This, by the definition, implies tht lim k n k = L Cuchy Sequence The definition of limit mkes eplicit use of the limit vlue L Therefore if we wnt to show the convergence of sequence by the definition, we need to first find the limit L nd then find suitble N for ech ɛ In mny cses, however, it is very hrd (or even impossible) to find the limit of convergent sequence For emple, we know the world record for meter dsh must hve limit set by the humn cpbility However, nobody knows ectly wht the humn cpbility is Proposition 7 gives specil cse tht we know the convergence of sequence without knowing the ctul limit vlue The following criterion provides method for the generl cse Definition A sequence { n } is clled Cuchy sequence if for ny ɛ >, there is N, such tht m, n > N implies n m < ɛ Theorem (Cuchy Criterion) A sequence converges if nd only if it is Cuchy sequence Proof Assume { n } converges to L For ny ɛ >, there is N, such tht n N implies n L < ɛ Then m, n > N = m L < ɛ, n L < ɛ = n m = ( n L) ( m L) n L + m L < ɛ + ɛ = ɛ The converse is much more difficult The convergence of Cuchy sequence { n } my be proved in the following steps The sequence is bounded By Bolzno-Weierstrss Theorem, the sequence hs convergent subsequence
LIMITS OF SEQUENCES 33 3 If Cuchy sequence hs subsequence converging to L, then the whole sequence converges to L Diligent reders my try to prove the first nd the third sttements The Bolzno-Weierstrss Theorem (Proposition 9) is very deep result tht touches the essentil difference between the rel nd rtionl numbers Emple 8 Consider the sequence n = + + 3 + + n If the sequence were convergent, the limit would be the sum of the infinite series n= n = + + 3 + + n + To show the convergence, we note tht for m > n, m n = = (n + ) + (n + ) + + m n(n + ) + (n + )(n + ) + + (m )m ( n ) ( + n + n + ) ( + + n + m ) m = n m Thus for ny ɛ, choose N = Then m > n > N implies ɛ n m < n < N = ɛ By the Cuchy criterion, the sequence { n } converges The ctul sum of the infinite series is n = π However, knowing the sum probbly still 6 n= gives you no clue how to prove the sequence converges Moreover, it took mthemticin gret effort to prove tht the sum of the series is trnscendentl number (not root of ny n 3 n= polynomil), nd we do not even know whether n= converges for ny p > np n= is trnscendentl or not Yet we know tht n5
34 LIMIT AND CONTINUITY Emple 9 We use the opposite of the Cuchy criterion to show tht the sequence n = + + 3 + + n diverges Note tht sequence is not Cuchy if there is ɛ >, such tht for ny N, we cn find m, n > N stisfying m n ɛ For ny n, we hve n n = n + + n + + + n n + n + + n = Choose ɛ = For ny N, we my find nturl number n > N Then we lso hve m = n > N nd m n = n n = ɛ Therefore, the sequence fils the Cuchy criterion nd diverges Eercises Find limit of sequence: n( n + n); n n ; 5 sin n! n ; 6 + + + + n, < ; 3 4 (n )! (n + )! ; 5 n n 6 n+ 3 n 3n+ ; 7 n n, < ; 8 n n Use the definition to prove the limit lim =, p > ; n np cos n lim n n = ; n + 3 3 lim n n n = ; π n 4 lim n n! = ; 5 lim n ( n n + ) = ; 6 lim n (n 3 (n ) 3 ) = ; 3 Determine convergence Provide rigorous reson
LIMITS OF SEQUENCES 35 ( )n n n 3 ; ( )n n n ; 3 ( )n n n ; 4 cos n n ; 9,, 3, 3, 3 4, 4 3,, n n +, n + n, ; + + 3 + + n ; 3 + 3 3 + ( )n+ n 3 ; +! + 3! + + n! ; 5 6 7 n n! ; n! n ; n + cos n n + ( ) n n + sin n ; 8,, 3, 3, 3 4, 4 3,, n n +, n + n, ; 3 n = +! + 3! + 4 3! + + n (n )! ; 4 5 3 5 (n ) ; 4 6 (n) 3 5 (n ) n! 4 Suppose lim n nb n = Cn you conclude tht either lim n n = or lim n b n =? 5 Prove tht if lim n n = L, then lim n n = L 6 Prove tht lim n n L = if nd only if lim n n = L 7 Prove tht if lim n n = L, then lim n c n = cl 8 For α > nd ny b >, prove tht the sequence defined by = b, n+ = ) ( n + αn stisfies n α for n nd is decresing Then prove tht the sequence converges to α Finlly discuss wht hppens when b < 9 Define the sequence { n } by = α nd Prove tht n+ = n +, n =,, 3, lim n = for ny rel number α, with α n + Let > nd n = n { n } is convergent Wht is the limit of this sequence? for ech positive integer n > Show tht the sequence
36 LIMIT AND CONTINUITY For α >, consider the sequence α, α + α, α + α + α, α + Assume the sequence converges, find the limit β Prove tht β is the upper bound of the sequence 3 Prove tht the sequence is incresing 4 Prove tht the sequence indeed converges to β For ny, b >, define sequence by Prove tht the sequence converges 3 The Fiboncci sequence α + =, = b, n = n + n,,, 3, 5, 8, 3,, 34, α + α, is defined by = = nd n+ = n + n Consider the sequence b n = n+ n Find the reltion between b n+ nd b n Assume the sequence {b n } converges, find the limit β 3 Use the reltion between b n+ nd b n to prove tht β is the upper bound of b k nd the lower bound of b k+ 4 Prove tht the subsequence {b k } is incresing nd the subsequence {b k+ } is decresing 5 Prove tht the sequence {b n } converges to β 4 Prove tht the sequences nd converge to e 5 Prove tht for n > k, we hve ( + ) n + n! +! The use Proposition 6 to show tht { ( + ) } n+ nd n ( n { ( ) } n re decresing, bounded, n ) + + ( ) ( ) ( k ) k! n n n e +! +! + + k! ( + k ) k
LIMITS OF FUNCTIONS 37 Finlly, prove ( lim + n! +! + + ) = e n! 6 If { n } is Cuchy, cn you conclude tht { n } is Cuchy? Why? Wht bout the converse? 7 Suppose tht the sequence { n } stisfies n+ n /n, n =,, 3, Prove tht { n } is convergent 8 If n n = sin(π/n), n =,, 3,, does the sequence { n } converge? Eplin 9 Suppose n is bounded sequence nd q < Prove tht the sequence n = + q + q + + n q n is Cuchy sequence nd therefore converges LIMITS OF FUNCTIONS Now we study limit of functions We will see tht it is close relted to limit of sequences induced by functions Concept of Limit nd Properties Consider the functions f() =, g() =, nd h() = sin s in Fig When pproches, f() = lso pproches, the bsolute vlue of g() = gets lrger nd does not pproch fied finite number Moreover, the vlue of h swings, such s ( ) ( ) h =, h ( ) nπ n + = π Therefore h() does not pproch ny one specific finite number We sw similr behviors { } for sequences when n pproches infinity The functions f(), g(), h() re comprble to, {( ) n n}, {( ) n } Only the first sequence converges This leds to the similr definition of limit for functions n Definition (Non-rigorous) A function f() converges to finite number L t = if f() pproches L s pproches We write lim f() = L
38 LIMIT AND CONTINUITY y y f() = y f() = / h() = sin Fig Grphs of f() =, g() = nd h() = sin Thus the limit lim = converges, nd the limits lim, lim sin diverge The limit of function hs properties similr to the limit of sequence Proposition (Arithmetic Rule) Suppose lim f() = L nd lim g() = K Then lim (f() + g()) = lim f + lim g, lim cf() = c lim f(), lim f()g() = lim f() lim g() Moreover, if lim g(), then f() lim f() lim g() = lim g() Proposition 3 If lim f() converges, then lim f() lso converges nd lim f() = lim f() On the other hnd, if lim f() =, then lim f() = Proposition 4 (Sndwich Rule) If f() g() h() nd lim f() = lim h() = L, then c c lim g() = L c
LIMITS OF FUNCTIONS 39 Proposition 5 (Order Rule) Suppose lim f() nd lim g() converge If f() g(), then lim f() lim g() If lim f() > lim g(), then f() > g() for sufficiently close to nd not equl to The convergence nd the limit vlue of sequence depends only on the terms with lrge inde n In other words, if finitely mny terms in sequence is dropped, dded, or modified, then the new sequence converges if nd only if the originl sequence converges Moreover, the limits of the two sequences re the sme Similrly, the convergence nd the limit vlue of function f() t depends only on the vlue of f() for close to nd not equl to Moreover, the function does not even need to be defined t in order for the limit to mke sense Therefore, if g() = f() for close but not equl to, then lim g() converges if nd only if lim g() converges Moreover, the limits of the two functions re the sme Emple We hve lim = nd lim c = c Then by the rithmetic rule, we hve lim = lim lim = =, lim c = c lim = c By the similr ide, we get lim c n = c n Then for ny polynomil p() = c n n + c n n + + c + c, we get lim p() = lim c n n + lim c n n + + lim c + lim c = c n n +c n n + +c +c = p() Moreover, rtionl function is the quotient of two polynomils r() = p() nd is defined t if q() q() Then further by the rithmetic rule, we hve lim r() = r() whenever r is defined t Emple The function 3 is not defined t = Yet the function converges t 3 lim = lim ( + + ) = + + = 3 Emple 3 By lim 3 + = nd the rithmetic rule, we get lim 3 + = lim ( + ) ( ( 3 + ) + 3 + + ) = (lim 3 + ) + lim 3 + + = 3 Emple 4 The limit lim sin cnnot be computed s follows lim sin = lim lim sin,
4 LIMIT AND CONTINUITY becuse the second limit on the right diverges However, if we use sin, nd the sndwich rule, then we get lim sin = Another useful property of the limit of function is the following The property cn be compred with Proposition 8 The sterisk indictes the sttement is not quite true Proposition 6 (Composition Rule*) If lim f() = b nd lim g(y) = c, then lim g(f()) = y b c For emple, the limit lim 3 + = is relly the composition of lim (+) = nd lim 3 y = y The composition rule cn be understood s follows Let y = f() nd z = g(y) Then z = g(f()) is the composition The ssumptions lim f() = b nd lim g(y) = c mens tht y b When the two implictions re combined, we get = y b, y b = z c = z c which mens lim g(f()) = c However, there is slight error in the eplntion bove, wht we relly hve re the following implictions, = y b, y b, y b = z c To mtch the right side of the first impliction to the left side of the second impliction, we must either modify the first impliction into, = y b, y b, which mens tht lim f() = b nd f() b for close nd not equl to, or modify the second impliction into y b = z c, which mens tht the limit lim y b g(y) = c lso includes the possibility of y = b It is not hrd to see tht the second modifiction mens lim y b g(y) = g(b), or g is continuous t b Emple 5 By lim sin = nd lim ( ) =, we get lim ( ) sin =