Bartle 4th Edition: Limits of Functions (PDF) - Mathematics

CHAPTER 4 LIMITS "Mathematical analysis" is generally understood to refer to that area of mathematics in which systematic use is made of various limiting concepts. In the preceding chapter we studied one of these basic limiting concepts: the l...

CHAPTER 4 LIMITS "Mathematical analysis" is generally understood to refer to that area of mathematics in which systematic use is made of various limiting concepts. In the preceding chapter we studied one of these basic limiting concepts: the limit of a sequence of real numbers. In this chapter we will encounter the notion of the limit of a function. The rudimentary notion of a limiting process emerged in the 1680s as Isaac Newton ( 1642-1 727) and Gottfried Leibniz ( 1 646- 1 7 1 6) struggled with the creation of the Calculus. Though each person's work was initially unknown to the other and their creative insights were quite different, both realized the need to formulate a notion of function and the idea of quantities being "close to" one another. Newton used the word "fluent" to denote a relationship between variables, and in his major work Principia in 1687 he discussed limits "to which they approach nearer than by any given difference, but never go beyond, nor in effect attain to, till the quantities are diminished in infinitum." Leibniz introduced the term "function" to indicate a quantity that depended on a variable, and he invented "infinitesimally small" numbers as a way of handling the concept of a limit. The term "function" soon became standard terminology, and Leibniz also introduced the term "calculus" for this new method of calculation. In 1 748, Leonhard Euler ( 1 707-1 783) published his two-volume treatise lntroduc­ tio in Analysin Infinitorum, in which he discussed power series, the exponential and logarithmic functions, the trigonometric functions, and many related topics. This was followed by Institutiones Calculi Differentia/is in 1 755 and the three-volume Institu­ tiones Calculi Integra/is in 1 768-1 770. These works remained the standard textbooks on calculus for many years. But the concept of limit was very intuitive and its looseness led to a number of problems. Verbal descriptions of the limit concept were proposed by other mathematicians of the era, but none was adequate to provide the basis for rigorous proofs. In 1 82 1 , Augustin-Louis Cauchy ( 1 789-1 857) published his lectures on analysis in his Cours d'Analyse, which set the standard for mathematical exposition for many years. He was concerned with rigor and in many ways raised the level of precision in mathematical discourse. He formulated definitions and presented arguments with greater care than his predecessors, but the concept of limit still remained elusive. In an early chapter he gave the following definition: If the successive values attributed to the same variable approach indefinitely a fixed value, such that they finally differ from it by as little as one wishes, this latter is called the limit of all the others. The final steps in formulating a precise definition of limit were taken by Karl Weierstrass ( 1 8 1 5-1 897). He insisted on precise language and rigorous proofs, and his definition of limit is the one we use today. 102  4. 1 LIMITS OF FUNCTIONS 103 Gottfried Leibniz Gottfried Wilhelm Leibniz ( 1 646-- 1 7 1 6) was born in Leipzig, Germany. He was six years old when his father, a professor of philosophy, died and left his son the key to his library and a life of books and learning. Leibniz entered the University of Leipzig at age 1 5, graduated at age 17, and received a Doctor of Law degree from the University of Altdorf four years later. He wrote on legal matters, but was more interested in philosophy. He also developed original theories about language and the nature of the universe. In 1 672, he went to Paris as a diplomat for four years. While there he began to study mathematics with the Dutch mathematician Christiaan Huygens. His travels to London to visit the Royal Academy further stimulated his interest in mathematics. His background in philosophy led him to very original, though not always rigorous, results. Unaware of Newtons's unpublished work, Leibniz published papers in the 1 680s that presented a method of finding areas that is known today as the Fundamental Theorem of Calculus. He coined the term "calculus" and invented the dy jdx and elongated S notations that are used today. Unfortunately, some followers of Newton accused Leibniz of plagiarism, resulting in a dispute that lasted until Leibniz's death. Their approaches to calculus were quite different and it is now evident that their discoveries were made independently. Leibniz is now renowned for his work in philosophy, but his mathematical fame rests on his creation of the calculus. Section 4.1 Limits of Functions In this section we will introduce the important notion of the limit of a function. The intuitive idea of the function/having a limit L at the point c is that the valuesf(x) are close to L when x is close to (but different from) c. But it is necessary to have a technical way of working with the idea of "close to" and this is accomplished in the B-8 definition given below. In order for the idea of the limit of a function f at a point c to be meaningful, it is necessary thatf be defined at points near c. It need not be defined at the point c, but it should be defined at enough points close to c to make the study interesting. This is the reason for the following definition. 4.1.1 Definition Let A c;;: R A point c E lR is a cluster point of A if for every 8 > 0 there exists at least one point x E A , x "I- c such that lx - ci 8. < This definition is rephrased in the language of neighborhoods as follows: A point c is a cluster point of the set A if every a-neighborhood V8 (c) = (c - 8, c + 8) of c contains at least one point of A distinct from c. Note The point c may or may not be a member of A, but even if it is in A , it is ignored when deciding whether it is a cluster point of A or not, since we explicitly require that there be points in V8 (c) n A distinct from c in order for c to be a cluster point of A. For example, i f A : = { 1 , 2 } , then the point 1 is not a cluster point of A , since choosing 8 : = ! gives a neighborhood of 1 that contains no points of A distinct from 1. The same is true for the point 2, so we see that A has no cluster points. 104 CHAPTER 4 LIMITS 4.1.2 Theorem A number c E lR is a cluster point of a subset A oflR if and only if there exists a sequence (an ) in A such that lim(an) = c and an -=/= c for all n E N. Proof. If c is a cluster point of A, then for any n E N the (1/n)-neighborhood V 1 ;n (c) contains at least one point an in A distinct from c. Then an E A , an -=/= c, and i an - c l implies lim (an) = c. 0 there exists K such that if n K, then an E V8 (c). Therefore the a-neighborhood V8 (c) of c contains the points aw for n K, which belong to A and are distinct from c. Q.E.D. The next examples emphasize that a cluster point of a set may or may not belong to the set. 4.1.3 Examples (a) For the open interval A 1 := (0, 1 ) , every point of the closed interval [0, 1 ] is a cluster point of A 1 Note that the points 0, 1 are cluster points of A ] , but do not belong to A 1 All the points of A 1 are cluster points of A 1. (b) A finite set has n o cluster points. (c) The infinite set N has no cluster points. (d) The set A4 : = { 1 / n : n E N} has only the point 0 as a cluster point. None of the points in A4 is a cluster point of A4. (e) If I : = [0, 1 ] , then the set A 5 : = I n Q consists of all the rational numbers in I. It follows from the Density Theorem 2.4.8 that every point in I is a cluster point of A5 D Having made this brief detour, we now return to the concept of the limit of a function at a cluster point of its domain. The Definition of the Limit We now state the precise definition of the limit of a function/ at a point c. It is important to note that in this definition, it is immaterial whether f is defined at c or not. In any case, we exclude c from consideration in the determination of the limit. 4.1.4 Definition Let A lR, a real number L is said to be a limit off at c if, given any 8 > 0, there exists a 8 > 0 such that if x E A and 0 l x - c l 8, then lf(x) - L l 8. < < < Remarks (a) Since the value of 8 usually depends on c: , we will sometimes write 8( ;) instead of 8 to emphasize this dependence. < (b) The inequality 0 l x - cl is equivalent to saying x -=/= c. If L is a limit of f at c, then we also say that f converges to L at c. We often write L = limf(x) or L = X..__.limf...., C K-----; C We also say that "f(x) approaches L as x approaches c." (But it should be noted that the points do not actually move anywhere.) The symbolism f(x) ---> L as x ---> c is also used sometimes to express the fact that f has limit L at c.  4. 1 LIMITS OF FUNCTIONS 105 If the limit of f at c does not exist, we say that f diverges at c. Our first result is that the value L of the limit is uniquely determined. This uniqueness is not part of the definition of limit, but must be deduced. 4.1.5 Theorem If f : A ----> lR and if c is a cluster point of A, then f can have only one limit at c. Proof. Suppose that numbers L and L' satisfy Definition 4. 1.4. For any c > 0, there exists a(c/2) > 0 such that if x E A and 0 < lx - cl < a(c/2) , then l f(x) - Ll < c/2. Also there exists a' (c/2) such that if x E A and 0 < lx - cl < a' (c/2) , then l f(x) - L' l < c/2. Now let a : = inf{a(c/2) , a' (c/2) }. Then if x E A and 0 < lx - cl < a, the Triangle Inequality implies that I L - L' l ::; IL - f(x) l + lf(x) - L' l < c/2 + c/2 = c. Since f; > 0 is arbitrary, we conclude that L - I! = 0, so that L = L'. Q.E.D. The definition of limit can be very nicely described in terms of neighborhoods. (See Figure 4. 1. 1.) We observe that because V8 (c) = (c - a, c + a) = { x : lx - cl < a }, the inequality 0 < lx - cl < a is equivalent to saying that x -1- c and x belongs to the a-neighborhood V8 (c) of c. Similarly, the inequality lf(x) - Ll < B is equivalent to saying that f(x) belongs to the £-neighborhood V, (L) of L. In this way, we obtain the following result. The reader should write out a detailed argument to establish the theorem. y L G i ven l'e (L) _...;.rc-to---- -+-------- ---------- x ---- c There exists Y,s(c) Figure 4.1.1 The limit off at c is L 4.1.6 Theorem Let f : A ----> lR and let c be a cluster point of A. Then the following statements are equivalen{ (i) limf(x) = L. (ii) Given any c-neighborhood V, (L) ofL, there exists a a-neighborhood V8 ( c ) of c such x c that if x -1- c is any point in V8 (c) n A , then f(x) belongs to V, (L). We now give some examples that illustrate how the definition of limit is applied. 4.1.7 Examples (a) lim b = b. x c To be more explicit, let.f (x) := b for all x E R We want to show that lim.f (x) = b. If c > 0 is given, we let a := 1. (In fact, any strictly positive a will serve the purpose.) Then if 106 CHAPTER 4 LIMITS 0 < lx - cl < 1 , we have I J(x) - bl = lb - bl = 0 < 8. Since 8 > 0 IS arbitrary, we conclude from Definition 4. 1.4 that Xlimf(x) -> C = b. lim-7 C x = c. (b) X--- Let g(x) : = x for all x E R If e > 0, we choose 8(8) := 8. Then if O < lx - cl < 8(c) , we have lg(x) - cl = lx - cl < 8. Since r; > 0 is arbitrary, we deduce that xlim c g = c. (c) lim x2 = c2. x c Let h(x) := x2 for all x E R We want to make the difference less than a preassigned c > 0 by taking x sufficiently close to c. To do so, we note that x2 - c2 = (x + c) (x - c). Moreover, if lx - cl < 1 , then lxl < l ei + 1 so that lx + cl 0 for an arbitrary choice of D > 0, we infer that lim h(x) = Xlim -+ C x2 = c 2. X -+ C 1. (d) X1 Ill. - = -1 I f C > 0. -+ C X c Let rp (x) : = 1 / x for x > 0 and let c > 0. To show that lim rp = I / c we wish to make x c the difference l rp (x) - 1 = 1 - 1 C X C less than a preassigned e > 0 by taking x sufficiently close to c > 0. We first note that 1 x - c = l __!_ex (c - x) l = __!_ex lx - cl for x > 0. It is useful to get an upper bound for the term 1 /(ex) that holds in some neighborhood of c. In particular, if lx - cl < ! c, then ! c < x < c (why?), so that 1 2 1 0 < - 0, we infer that D X---+ C x3 - 4 = -4 (e) x lim2 x2 + 1 5 -- Let lfr(x) : = (x3 - 4)j(x2 + 1) for x E JR. Then a little algebraic manipulation gives us l 5x3 - 4x2 - 24 1 I lfr( x) - 5 1 -- 5 (x2 + 1) =l 5x3 + 6x + 12 1 · l x - 2 1. 5(x2 + J) To get a bound on the coefficient of l x - 21 , we restrict x by the condition 1 < x < 3. For x in this interval, we have 5x2 + 6x + 12 ::; 5 · 3 2 + 6 · 3 + 12 = 7 5 and 5(x2 + 1) 5(1 + I) = 1 0, so that 2: I lfr(x) - 45 1 TO75 l x - 2 1 215 l x - 2 1. :S = Now for given B > 0, we choose 2 8 (£ ) := inf { 1, £ } · 15 Then if O < l x - 2 1 < 8(c), we have l lfr (x) - (4/5) 1 ::; (15/2)l x - 2 1 < B. Since > 0 is c; arbitrary, the assertion is proved. 0 Sequential Criterion for Limits ___________________ 3 The following important formulation of limit of a function is in terms of limits of sequences. This characterization permits the theory of Chapter to be applied to the study of limits of functions. 4.1.8 Theorem (Sequential Criterion) Let f : A lR and let c be a cluster point of A. ---+ Then the following are equivalent. (i) limf = L. X---+ C (f(xn)) (ii) For every sequence sequence converges to L. (xn) in A that converges to c such that =/= c for all n E N, the Xn (i) =? (ii). Assume c, and suppose (xn) is a sequence in A with (xn) = c and n =/= c for allf hasn. WelimitmustL atprove the sequence (f(xn)) converges to L. Proof. lim Let B > 0 be given. Then by Definition 4.1.4, therethatexists X 8 > 0 such that if x E A satisfies 108 CHAPTER 4 LIMITS 0 < lx - cl < 8, then f(x) satisfies l f(x) - Ll < e. We now apply the definition of convergent sequence for the given 8 to obtain a natural number K(8) such that if n > K(8) then l xn - cl < 8. But for each such Xn we have l f(xn) - Ll < e. Thus if n > K(8) , then l f(xn) - Ll < e. Therefore, the sequence (f(xn)) converges to L. (ii) =? (i). [The proof is a contrapositive argument.] If (i) is not true, then there exists an eo-neighborhood Vc0 (L) such that no matter what 8-neighborhood of c we pick, there will be at least one number x8 in A n Va (c) with xa -=1- c such thatf(x8) ¢:. Vc0 (L). Hence for every n E N, the ( 1 /n)-neighborhood of c contains a number Xn such that and Xn E A , but such that for all n E N. We conclude that the sequence (xn ) in A\ { c} converges to c, but the sequence (!( xn ) ) does not converge to L. Therefore we have shown that if (i) is not true, then (ii) is not true. We conclude that (ii) implies (i). Q.E.D. We shall see in the next section that many of the basic limit properties of functions can be established by using corresponding properties for convergent sequences. For example, we know from our work with sequences that if (xn ) is any sequence that converges to a number c, then ( x ) converges to c2. Therefore, by the sequential criterion, we can conclude that the function h(x) := x2 has limit lim h(x) = c2. X-->C Divergence Criteria -------­ It is often important to be able to show (i) that a certain number is not the limit of a function at a point, or (ii) that the function does not have a limit at a point. The following result is a consequence of (the proof of) Theorem 4. 1.8. We leave the details of its proof as an important exercise. 4.1.9 Divergence Criteria Let A c::; R let f : A lR and let c E lR be a cluster ---+ point of A. (a) IfL E R thenfdoes not have limit L at c ifand only if there exists a sequence (xn ) in A with Xn -=1- c for all n E N such that the sequence (xn ) converges to c but the sequence (!( xn) ) does not converge to L. (b) The function f does not have a limit at c if and only if there exists a sequence (xn ) in A with Xn -=1- c for all n E N such that the sequence (xn ) converges to c but the sequence (f(xn)) does not converge in R We now give some applications of this result to show how it can be used. 4.1.10 Examples (a) lim ( 1 /x) does not exist in R X-->0 As in Example 4. 1.7(d), let rp(x) := 1 /x for x > 0. However, here we consider c = 0. The argument given in Example 4. 1.7(d) breaks down if c = 0 since we cannot obtain a bound such as that in (2) of that example. Indeed, if we take the sequence (xn ) with Xn := 1 /n for n E N, then lim (xn) = 0, but rp(xn ) = 1 / ( 1 /n) = n. As we know, the sequence (rp(xn) ) = (n) is not convergent in R since it is not bounded. Hence, by Theorem 4. 1.9(b), lim ( 1 /x) does not exist in R X-->0 (b) lim sgn(x) does not exist. X-->0  4. 1 LIMITS OF FUNCTIONS 109 Let the signum function sgn be defined by sgn(x) := { +10 forfor xx => 0,0, I for x < 0. Note that sgn(x) = x/l x l for x i= 0. (See Figure 4.1. 2. ) We shall show that sgn does not - have Iim(x,)a limit = 0,atbutx =such0. Wethatshall do thisdoes (sgn(x,)) by showing that there is a sequence (x11) such that not converge. ------1 - 1 Figure 4.1.2 The signum function Indeed, let Xn := ( -1 ) " /n for n E N so that Iim(xn ) = 0. However, since sgn(xn ) = ( -1 ) " for n E N , it follows from Example 3. 4. 6(a) that (sgn(xn )) does not converge. Therefore lim sgn(x) doest not exist. x-->O (c) lim sin(l/x) does not exist in R Let ( 1 /x) for xtwoi= 0.sequences (See Figure 4.1. 3. ) We shall show that g does not have a limit atg (x)= :0,= bysinexhibiting X-->0 (x,) and (y11) with Xn "!= 0 and Yn "!= 0 for all n E N and such that lim(xn ) = 0 and lim(y11) = 0, but such that lim(g(xn )) "!= lim(g(y11)). c In view of Theorem 4.1.9 this implies that lim g cannot exist. (Explain why. ) x-->0 Figure 4.1.3 The function g(x) = sin ( l /x) (x # 0) Indeed, we recall from calculus that sin t = 0 if t = mr for n E Z, and that sin t = + 1 if t = ! n + 2;r n for n E Z. Now let Xn := 1/mr for n E N; then lim(xn ) = 0 and g(xn ) :== si(!nnnn =2nnr 0 for1 forall nnEEN;N, then so thatlim(y11)lim(g(x = 0 n))and= g0.(y11)On= thesin (!other hand, let Yn n + 2n n ) = l for all n E N, so that Iim(g(y11)) = 1. We conclude that Iim sin(l/x) does not exist. + D X-->0 t In order to have some interesting applications in this and later examples, we shall make use of well-known properties of trigonometric and exponential functions that will be established in Chapter 8. 1 10 CHAPTER 4 LIMITS Exercises for Section 4.1 1. Determine a condition on lx - I I that will assure that: (a) l x2 - I I < , (b) l x2 - I I < l / I 0- 3 , (c) l x2 - l l < l /n for a given n E N , (d) l x3 - l i < I /n for a given n E N. 2. Determine a condition on l x - 41 that will assure that: (a) I JX - 21 < , (b) l v'X - 21 < 1 0 - 2. 3. Let c be a cluster point of A lR and let f : A --+ R Prove that limf(x) = L if and only if x c lim f(x) - L l = 0. x........, c l 4. Let f : = lR --+ lR and let c E R Show that limf(x) = L if and only if Iimf(x + c ) = L. x-c x-0 5. Let I := (0, a) where a > 0, and let g(x) := x2 for x E I. For any points x, c E I, show that l g (x) - c2 1 ::; 2alx - cl. Use this inequality to prove that x-c lim x2 = c2 for any c E I. 6. Let I be an interval in R letf : I --+ R and let c E I. Suppose there exist constants K and L such limf(x) = L. that l f(x) - L l ::; K l x - cl for x E I. Show that x-c 7. Show that lim x3 = c3 for any c E R x-c 8. Show that lim v'x = JC for any c > 0. x c 9. Use either the e-8 definition of limit or the Sequential Criterion for limits, to establish the following limits.. I. X 1 (a) hm -- = -1, (b ) hm-- =- x 2 ] - X x I I + X 2 ' xz x2 - x + I (c) lim0 - = 0, (d) lim x- 1 X I x l X + ] 1 0. Use the definition of limit to show that (a) lim (x2 + 4x ) = 1 2, x+5 (b) lim = 4. x 2 x - I 2x + 3 1 1. Use the definition of limit to prove the following. 2x + 3 x2 - 3x (a) lim -__ = 3, (b) lim6 = 2. x 3 4x - 9 x X + 3 1 2. Show that the following limits do not exist. (a) 2 (x > 0) , I (b) limo - (x > 0) , x JX (c) lim (x + sgn(x) ) , (d) limo sin ( l /x2 ). X----> 0 x 1 3. Suppose the function f : lR --+ lR has limit L at 0, and let a > 0. If g : lR --+ lR is defined by g(x) := f(ax) for x E R show that limo g(x) = L. x J 4. Let c E lR and let f : lR --+ lR be such that lim ( ! ( x) ) 2 = L. x c (a) Show that if L = 0, then x-c limf(x) = 0. (b) Show by example that if L # 0, then f may not have a limit at c. 1 5. Let f : lR --+ lR be defined by setting f(x) := x if x is rational, and f(x) = 0 if x is irrational. (a) Show that f has a limit at x = 0. (b) Use a sequential argument to show that if c # 0, then f does not have a limit at c. 16. Letf : lR --+ R let I be an open interval in R and let c E I. Ifj1 is the restriction off to I, show that f1 has a limit at c if and only if f has a limit at c, and that the limits are equal. 1 7. Let f : lR --+ R let J be a closed interval in R and let c E J. If fz is the restriction off to J, show that ifjhas a limit at c thenfz has a limit at c. Show by example that it does not follow that if f2 has a limit at c, then f has a limit at c.  4.2 LIMIT THEOREMS 111 Section 4.2 Limit Theorems We parallel are shall nowto obtain the limitresults that areestablished theorems useful in incalculating Section 3.limits of functions.InThese 2 for sequences. fact, inresults most cases these results Alternatively, the can results be in proved this by section using can Theorem be proved 4.1. by 8 and using results s-8 from arguments Section that are 3. 2. very similar to the ones employed in Section 3. 2. thatfis bounded onLeta neighborhood 4.2.1 Definition A IR, letf : A IR, and let c E IR be a cluster point of A. We say of c if there exists a 8-neighborhood V8 (c) of c and a -+ constant M > 0 such that we have l f(x) l :::; M for all x E A n V8 (c). 4.2.2 Theorem IfA IR andf : A -+ IR has a limit at c E IR, thenfis bounded on some neighborhood of c. If L := limf, then for = 1 , there exists 8 > 0 such that if 0 < lx - c l < 8, then l f(x) - Ll < 1 ; hence (by Corollary 2. 2. 4 (a)), Proof. c: X-----+ C l f(x) l - I L l :::; l f(x) - Ll < 1. Therefore, if then x E A n V0 (c) , x "/:- c, If we take while if we take := sup{ It follows that if l f(x) l :::; I Ll + 1. c f- A, then M = I Ll + 1 , This shows thatfis bounded on the neighborhood of cEA M I f( c) I , I L l + 1 }. x E A n V8 (c) , l f(x) l :::; M. V8 (c) c. Q.E.D. The next definition is similar to the definition for sums, differences, products, and quotients of sequences given in Section 3. 2. 4.2.3 Definition Let A IR and letf and g be functions defined on A to R We define the sumf + g, the differencef - g, and the productfg on A to IR to be the functions given by (f + g) (x) : = f(x) + g(x) , (f - g) (x) : = f(x) - g(x) , (fg) (x) : = f(x)g(x) for all x E A. Further, if b E IR, we define the multiple bf to be the function given by (bf) (x) : = bf(x) x EA. forall Finally, if h( x) "/:- 0 x E A, for we define the quotient f/ h to be the function given by f h (x) : = () f(x) h(x) for all X EA. R 4.2.4 Theorem Let A IR, let f and g be functions on A to IR, and let c E IR be a cluster point of A. Further, let b E (a) If limf = L and g = M, then: lim lim X----+ C X--+C !im (f + g) = L + M, (f - g) = L - M, lim (fg) = LM, lim (bf) = bL. X -> C X-----+ C IR, if h(x) "/:- 0 for all x E A, and if lim h = H "/:- 0, then X-+C X->C (b) If h : A -+ lim (£h) = Hl:_. X->C X->C 1 12 CHAPTER 4 LIMITS One proof of this theorem is exactly similar to that of Theorem 3.2.3. Alterna­ tively, it can be proved by making use of Theorems 3.2.3 and 4. 1.8. For example, let (xn) be Proof. any sequence in A such that Xn -=1- c for E N, and c = lim(xn) · It follows from Theorem 4. 1.8 that n lim(f(xn)) = L, lim(g(xn)) = M. On the other hand, Definition 4.2.3 implies that for E N. Therefore an application of Theorem 3.2.3 yields n lim((fg)(xn)) = lim(f(xn )g(xn )) = [lim(f (xn )) ] [lim(g(xn)) ] = LM Consequently, it follows from Theorem 4. 1. 8 that lim (fg) = lim((fg)(xn)) = LM. The other parts of this theorem are proved in a similar manner. We leave the details to X-----+ C the reader. Q.E.D. fz ,.. , fn be functions on A to JR., and let c be a cluster point of A.LetIf L cA :=JR.,limf Remark 0, then it follows from this result that since hm-x = limx = ­cI · X--+C X-----+ C. l x c 1 x c -- (b) lim (x2 + I ) (x3 - 4 ) = 20. x It follows 2 from Theorem 4.2.4 that ! (x2 + 1 ) (x3 - 4 ) = (! (x2 + 1 ) (! (x3 - 4 ) = 5. 4 = 20. ) ) (c) H-Tz c: ) · =  4.2 LIMIT THEOREMS 1 13 If we apply Theorem 4. 2.4(b ), we have lim-- x3 4 = '-x lim--- 2 (x3 --4) = -4 x 2 x2 + l x lim2 (x--,---,- 2 + 1)..,.. 5 _ Note that since the limit in the denominator [i. e. , l x i m2 (x 2 + 1) = 5] is not equal to 0, then Theorem 4.2.24(b) is applicable. (d) lim = -- = -. x -4 4 x 2 3x - 6 3 2 4. 2.4If(b)wetoletf(x) evaluate:=x limx2 (!(- 4x)and/h(h(x) := 3x - 6 for x E IR, then we cannot use Theorem x)) because H = X----7lim2 h(x) X--+lim2 (3x - 6) = 3 · 2 - 6 = 0. However, if x i= 2, then it follows that x2 - 4 _- (x + 2)(x - 2) -_ (x + 2) · 3x - 6 3(x - 2) 3 Therefore we have x2 - 4 x Jim2 3X - 6 = x lim 2 3 (x + 2) = 3 (lim x-+2 x + 2) = 3. Note that the function g(x) = (x2 - 4)/(3x - 6) has a limit at x = 2 even though it is not defined there. (e) lim does not exist in R X-+Of course X lim 1 = 1 and := x--+0 lim x = 0. However, since H = 0, we cannot use x--+0 O H Theorem 4.2.4(b) to evaluate X----+lim0 (1/x). In fact, as was seen in Example 4.1.10(a), the function cp(x) Theorem 4. 2. 2=sincel/xthedoesfunction not havecp(x)a limit = l/xatisxnot= 0.bounded This conclusion also follows on a neighborhood of xfrom = 0. (f) If p is a polynomial function, then limp(x) = p(c). X--tC Let p be a polynomial function on IR so that p(x) = anxn + an_ 1 xn-k l + ·k· ·that+ a1x + a0 for all x E R It follows from Theoremn 4. 2. 4 and the fact that X---+limC x = c limp(x) = X--+C X-+C lim [an + an- J X - l + · + a1x + ao] · · = X--tlimC (anxn ) + X--+C lim (an- J Xn - l ) + · · · + X----l-limC (a1x) + X---+limao C = p(c). Hence X----limp(x) -+ C - p(c) for any polynomial function p. = (g) If p and q are polynomial functions on IR and if q(c) i= 0, then lim p(x) x-+c q(x) = q(c) · p(c) Since q(x) is a polynomial function, it follows from a theorem in algebra that there are at most a finite number of real numbers [the real zeroes of q(x)] such that q ( ) = 0 and such that if x f_ { } then q(x) i= 0. Hence, if x ¢'- { } we can define a 1 ,... , am a1 p(x). a 1 ,... , am , a 1 ,... , am , r(x) := q(x) 1 14 CHAPTER 4 LIMITS If cis not a zero of q(x), then q( c) -1 0, and it follows from part (f) that xlim c q(x) = q( c) -1 0. Therefore we can apply Theorem 4.2.4(b) to conclude that p(x) p(x) p(c) 0 q(x) q(c) · lim = = x c q(x) lim x c The next result is a direct analogue of Theorem 3. 2. 6. 4.2.6 Theorem Let A IR, and let c E IR be a cluster point of A. If a :;_ f(x) :;. b for all x E A , x # c, and iflim lim f exists, then a :;_ f :;_ b. Proof. Indeed, if L = limf, then it follows from Theorem 4.1. 8 that if (xn ) is any X -+ C X-+C sequence of real numbers uch that c -1 Xn E A for all n E N and if the sequence (xn) converges to c, then the sequence (f(xn)) converges to L. Since a :;_ f(xn ) :;_ b for all n E N, it follows from Theorem 3. 2. 6 that a :;_ L :;_ b. Q.E.D. We now state an analogue of the Squeeze Theorem 3. 2. 7. We leave its proof to the reader. 4.2.7 Squeeze Theorem Let A IR, and let c E IR be a cluster point of A. If f(x) :;_ g(x) :;_ h(x) for all x E A, x -1 c, and if limf = L = X -+ C lim h, then lim g = L. X----t C X-+C 4.2.8 Examples (a) limo x3 1 2 = 0 (x > 0). x Let f(x) := x3 1 2 for x > 0. Since the inequality x < x 1 12 :;_ 1 holds for 0 < x :;_ 1 (why?), it follows that x2 :;_ f(x) = x312 :;_ x for 0 < x 1. Since :;_ lim x2 = 0 and lim x = 0 ' it follows from the Squeeze Theorem 4.2.7 that x limo x312 = 0. x-+0 x-+0 (b) limo sin x = 0. It will be proved later (see Theorem 8. 4. 8 ), that x - x :;_ sin x x for all x 2': 0. :;_ Since lim ( ±x) = 0, it follows from the Squeeze Theorem that lim sin x = 0. (c) limo cos x = 1. X -+ 0 X -+ 0 It will be proved later (see Theorem 8. 4. 8 ) that x (1) 1 - x2 :;_ cos x :;_ 1 for all x E R Since lim ( 1 - x2 ) = 1 , it follows from the Squeeze Theorem that lim cosx = 1. (cosx - 1 ) = 0. X-+0 X--+0 (d) lim x O X  4.2 LIMIT THEOREMS 1 15 We cannot use Theorem 4.2.4(b) to evaluate this limit. (Why not?) However, it follows from the inequality ( 1 ) in part (c) that - x ::; (cos x - I ) I x ::; 0 for x > 0 and that 0 ::; (cos x - 1 ) I x ::; - x for x < 0. Now Ietf(x) : = -xl2 for x 0 andf(x) : = 0 for x < 0, and let h(x) : = 0 for x 0 and h(x) := -xl2 for x < 0. Then we have f(x) ::; (cos x - I ) lx ::; h(x) for x -/:- 0. Since it is readily seen that lim/ = 0 = lim h, it follows from the Squeeze Theorem that ( ) X--+0 X--+0 I'! ill ( COS X - I ) l X = 0. 0 sin x. lim = 1 X--+0 X Again we cannot use Theorem 4.2.4(b) to evaluate this limit. However, it will be proved later (see Theorem 8.4.8) that x - * x3 ::; sin x ::; x for x 0 and that x ::; sin x ::; x - * x3 for x ::; 0. Therefore it follows (why?) that J - * x2 :S (sin x) lx :S l for all x -.f:- 0. But since lim ( 1 - * x2 ) = I - * · lim x2 = I , we infer from the Squeeze Theorem that x--+0 x--+0 lim (sin x) I x = 1. x--+0 (f) lim (x sin( l lx) ) = 0. X --+ 0 Let f(x) = x sin ( l lx) for x -/:- 0. Since - 1 ::; sin z ::; 1 for all z E JR, we have the inequality - lxl :S f(x) = x sin( 1 lx) ::; lxl for all x E lR, x -/:- 0. Since limo lxl = 0, it follows from the Squeeze Theorem that lim/ = 0. x x- o For a graph, see Figure 5. 1.3 or the cover of this book. D There are results that are parallel to Theorems 3.2.9 and 3.2. 1 0; however, we will leave them as exercises. We conclude this section with a result that is, in some sense, a partial converse to Theorem 4.2.6. 4.2.9 Theorem Let A 0). --­ X--+0 X x I X - 1 )1 + 2x - JI + 3x 3. Find lim where x > 0. x o x + 2x2 4. Prove that Jim cos( l /x) does not exist but that lim x cos ( l /x) = 0. x--+0 x--+0 5. Letf, g be defined on A lR to lR, and let c be a cluster point of A. Suppose thatfis bounded on a neighborhood of c and that x--+c lim g = 0. Prove that x--+c limfg = 0. 6. Use the definition of the limit to prove the first assertion in Theorem 4.2.4(a). 7. Use the sequential formulation of the limit to prove Theorem 4.2.4(b). 8. Let n E N be such that n ::0: 3. Derive the inequality -x2 ::; xn ::; x2 for - 1 < x < I. Then use the fact that lim x2 = 0 to show that limO x" = 0. X--+0 X--t 9. Let f, g be defined on A to lR and let c be a cluster point of A. (a) Show that if both x- lim (f + g) exist, then x c limf and x c lim g exists. (b) If limf and X--+C limfg exist, does it follow that lim g exists? X--+C X--+C 1 0. Give examples of functionsfand g such thatfand g do not have limits at a point c, but such that both f + g and fg have limits at c. 1 1. Determine whether the following limits exist in JR. lim sin ( l /x2 ) (x 7" 0) , (a) x--+ lim x sin ( 1 /x2 ) (x 7" 0) , (b) x--+0 (d) lim y'X sin ( l /x2 ) (x > O). 0 lim sgn sin ( 1 /x) (x T'o O) , (c) X--+0 X--+0 1 2. Let f : lR ---+ lR be such that f(x + y) = f(x) + f(y) for all x, y in JR. Assume that limf = L x o exists. Prove that L = 0, and then prove thatfhas a limit at every point c E JR. [Hint: First note that f(2x) = f(x) + f(x) = 2f(x) for x E JR. Also note that f(x) = f(x - c) + f(c) for x, c in JR.] 1 3. Functions / and g are defined on R by f(x) := x + 1 and g (x) : = 2 if x 7" 1 and g( l ) := 0. (a) Find lim g (f(x)) and compare with the value of g (lim f(x)). X--+ 1 X--+ J (b) Find lim.f(g (x)) and compare with the value of f(limJ g (x)). 14. Let A lR, let f : A ---+ lR and let c E lR be a cluster point of A. If limf exists, and if If I denotes X--+ 1 X--+ the function defined for x E A by lfl (x) := l f(x) l, prove that 1 lr l l = l f. 1 5. Let A lR, let f : A ---+ lR, and let c E lR be a cluster point of A. In addition, suppose that f(x) ::0: 0 for all x E A, and let JJ be the function defined for x E A by ( Jl) (x) : = Jl(X). If 1 / exists, prove that l T,. J.i = · Section 4.3 Some Extensions of the Limit Conceptt In this section, we shall present three types of extensions of the notion of a limit of a function that often occur. Since all the ideas here are closely parallel to ones we have already encountered, this section can be read easily. t This section can be largely omitted on a first reading of this chapter.  4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 1 17 One-Sided Limits ----- There are times when a function f may not possess a limit at a point c, yet a limit does exist when the function is restricted to an interval on one side of the cluster point c. For example, the signum function considered in Example 4. 1. 1 O(b ), and illustra­ ted in Figure 4. 1.2, has no limit at c = 0. However, if we restrict the signum function to the interval (0, oo ), the resulting function has a limit of 1 at c = 0. Similarly, if we restrict the signum function to the interval ( - oo , 0), the resulting function has a limit of - 1 at c = 0. These are elementary examples of right-hand and left-hand limits at c = 0. 4.3.1 Definition Let A E IR and let f : A ---+ R (i) If c E IR is a cluster point of the set A n ( c, ) = { x E A: x > c }, then we say that oo L E IR is a right-hand limit off at c and we write lim f = L or X---t c+ lim f(x) = L X---t c+ if given any ;; > 0 there exists a 8 = 8(;;) > 0 such that for all x E A with < < 0 x - c 8, then lf(x) - Ll c. < < (ii) If c E IR is a cluster point of the set A n ( - oo , c) = { x E A: x c }, then we say that L E IR is a left-hand limit of f at c and we write lim f = L or lim f(x) X---t C- = L X----> C- if given any c > 0 there exists a 8 > 0 such that for all x E A with 0 then l f(x) - L l < 1:. < c x < 8, - Notes ( l ) The limits X----+ lim-+ C - f are called one-sided limits offat c. It is possible limc+ f and X---- that neither one-sided limit may exist. Also, one of them may exist without the other existing. Similarly, as is the case for f(x) := sgn (x) at c = 0, they may both exist and be different. (2) If A is an interval with left endpoint c, then it is readily seen thatf : A ---+ IR has a limit at c if and only if it has a right-hand limit at c. Moreover, in this case the limit X----+limf C and the right-hand limit x----+ limc+ f are equal. (A similar situation occurs for the left-hand limit when A is an interval with right endpoint c.) The reader can show thatfcan have only one right-hand (respectively, left-hand) limit at a point. There are results analogous to those established in Sections 4. 1 and 4.2 for two­ sided limits. In particular, the existence of one-sided limits can be reduced to sequential considerations. 4.3.2 Theorem Let A c for all n E N, the sequence (f( xn ) ) converges to L. 118 CHAPTER 4 LIMITS y -----o.;;: _ _Y__" Figure 4.3.2 Graph of Figure 4.3.1 Graph of h( x) = 1 / ( e 1 fx + 1 ) (x f= 0) g(x) = e 1 fx (x f= 0) We leave the proof of this result (and the formulation and proof of the analogous result for left-hand limits) to the reader. We will not take the space to write out the formulations of the one-sided version of the other results in Sections 4. 1 and 4.2. The following result relates the notion of the limit of a function to one-sided limits. We leave its proof as an exercise. 4.3.3 Theorem Let A R let f : A R and let c E IR be a cluster point of both ----+ of the sets A n (c, oo) and A n ( - oo, c). Then lim f = L if and only if lim f = L = lim f. x c X--l-C + X----+ C- 4.3.4 Examples (a) Let f(x) := sgn (x). We have seen in Example 4. 1. 1 O(b) that sgn does not have a limit at 0. It is clear that lim sgn (x) = + 1 and that lim sgn (x) = - 1. Since these one-sided limits are different, it x O + x O- also follows from Theorem 4.3.3 that sgn(x) does not have a limit at 0. (b) Let g (x) := e 1 fx for x -=? 0. (See Figure 4.3. 1.) We first show that g does not have a finite right-hand limit at c = 0 since it is not bounded on any right-hand neighborhood (0, 8) of 0. We shall make use of the inequality (I) 0 < t < e 1 for t > 0, which will be proved later (see Corollary 8.3.3). It follows from ( 1 ) that if x > 0, then 0 < I lx < e 1 1 x. Hence, if we take Xn = l in, then g(xn) > n for all n E N. Therefore lim e 1 1x does not exist in R x D +However, lim e 1 I x = 0. Indeed, if x < 0 and we take t = - I I x in ( 1 ) we obtain x----+ 0 - 0 < - I Ix < e - l /x. Since x < 0, this implies that 0 < e 1 fx < -x for all x < O. lt follows from this inequality that lim0 e 1 fx = 0. x----+ - (c) Let h (x) : = 1 I ( e 1 1x + 1 ) for x -=1 0. (See Figure 4.3.2.) We have seen in part (b) that 0 < l lx < e 1 fx for x > 0, whence 1 1 O < 1 /x < < x, e + 1 e l /x which implies that lim+ h = 0. x O  4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 1 19 Since we have seen in part (b) that lim e 11x = 0, it follows from the analogue of X---+ 0 - Theorem 4.2.4(b) for left-hand limits that r ( I ) = 1 x = I = I· x W- e l/ + I.!.. x lim e 11 + I 0 + 1 ---+ 0 - x Note that for this function, both one-sided limits exist in IR, but they are unequal. 0 Infinite Limits The functionf(x) := 1 j x2 for x -=/= 0 (see Figure 4.3.3) is not bounded on a neighborhood of 0, so it cannot have a limit in the sense of Definition 4. 1.4. While the symbols oo(= +oo) and -oo do not represent real numbers, it is sometimes useful to be able to say that "f(x) = l jx2 tends to oo as x -+ 0." This use of ±oo will not cause any difficulties, provided we exercise caution and never interpret oo or -oo as being real numbers. Figure 4.3.3 Graph of f(x) = l jx2 (x f= 0) Figure 4.3.4 Graph of g(x) = 1 /x (x f= 0) 4.3.5 Definition Let A R let f : A -+ IR, and let c E IR be a cluster point of A. (i) We say that f tends to oo as x -+ c, and write limC f = oo, X---+ if for every a E IR there exists 8 = 8(a) > 0 such that for all x E A with < < 0 lx - c l 8, then f(x) > a. (ii) We say that f tends to -oo as x -+ c , and write limC f = -oo, X---+ f3 - 0if 0 such that for all x E < {3. A with 4.3.6 Examples (a) lim ( I /x2 ) oo. = x o For, if a > 0 is given, let 8 := 1 j fo. It follows that if 0 that l jx2 > a. < lxl < 8, then x2 < I fa so (b) Let g(x) : = 1 /x for x -=/= 0. (See Figure 4.3.4.) 120 CHAPTER 4 LIMITS The function g does not tend to either oo or - oo as x ----+ 0. For, if a > 0 then g(x) < a for all x < 0, so that g does not tend to oo as x ----+ 0. Similarly, if f3 < 0 then g(x) > f3 for all x > 0, so that g does not tend to -oo as x ----+ 0. D While many of the results in Sections 4. 1 and 4.2 have extensions to this limiting notion, not all of them do since ±oo are not real numbers. The following result is an analogue of the Squeeze Theorem 4.2.7. (See also Theorem 3.6.4.) 4.3.7 Theorem Let A IR, let f, g : A ----+ IR, and let c E IR be a cluster point of A. Suppose that f(x) :::; g(x) for all x E A, x -=/= c. (a) If limf = oo, then lim g = oo. X----> C X----> C (b) If lim g = - oo, then limf = - oo. X----+ C X-H' Proof. (a) If limf = oo and a E IR is given, then there exists 8(a) > 0 such that if X----+ C 0 < l x - cl < 8(a) and x E A, then f(x) > a. But since f(x) :::; g(x ) for all x E A , x -=/= c, it follows that if 0 < lx - cl < 8(a) and x E A, then g(x) > a. Therefore X-"C lim g = oo. The proof of (b) is similar. Q.E.D. The function g(x) = 1 /x considered in Example 4.3.6(b) suggests that it might be useful to consider one-sided infinite limits. We will define only right-hand infinite limits. 4.3.8 Definition Let A IR and let f : A ----+ R If c E IR is a cluster point of the set An (c, oo) = {x E A : x > c}, then we say that f tends to oo [respectively, - oo] as x ----+ c+ , and we write [ lim f = oo respectively, lim f = -oo , x ----+ c+ x ----+ c+ ] if for every a E IR there is 8 = 8 ( a) > 0 such that for all x E A with 0 < x - c < 8, then f(x) > a [respectively, f(x) < a]. 4.3.9 Examples (a) Let g (x) : = 1 /x for x -=/= 0. We have noted in Example 4.3.6(b) that lim g does not exist. However, it is an easy exercise to show that x ----+ 0 lim ( 1 /x) = oo and lim ( 1 /x) = -oo. X----+ 0 + X----+ 0- (b) It was seen in Example 4.3.4(b) that the function g(x) := e 1 1 x for x -=/= 0 is not bounded on any interval (0, 8 ) , 8 > 0. Hence the right-hand limit of e 1 1x as x ----+ 0+ does not exist in the sense of Definition 4.3. 1 (i). However, since 1 /x < e 1 1 for x > 0 , x it is readily seen that lim e 1 1 = oo in the sense of Definition 4.3.8. x D x O+ Limits at Infinity It is also desirable to define the notion of the limit of a function as x ----+ oo. The definition as x ----+ - oo is similar.  4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 121 4.3.10 Definition Let A C:: IR and let f : A ---+ R Suppose that (a, oo) C:: A for some a E R We say that L E IR is a limit of f as x ---+ oo, and write Jim f(x) = L ' lim f = L or X----+ 00 X---+ 00 if given any D > 0 there exists K = K( ;) > a such that for any x > K, then lf(x) - Ll < B. The reader should note the close resemblance between 4.3. 10 and the definition of a limit of a sequence. We leave it to the reader to show that the limits of f as x ---+ ±oo are unique whenever they exist. We also have sequential criteria for these limits; we shall only state the criterion as x ---+ oo. This uses the notion of the limit of a properly divergent sequence (see Definition 3.6. 1 ). 4.3.11 Theorem Let A C:: R let f : A IR, and suppose that (a, oo) C:: A for some ---+ a E R Then the following statements are equivalent: (i) L = lim f. X-->00 (ii) For every sequence (x11) in A n (a, oo) such that lim ( xn ) = oo, the sequence (f(xn ) ) converges to L. We leave it to the reader to prove this theorem and to formulate and prove the companion result concerning the limit as x ---+ -oo. 4.3.12 Examples (a) Let g(x) : = 1 I x for x -j. 0. It is an elementary exercise to show that x---+ limoo ( I I x) = 0 = x---+lim - oo ( 1 I x). (See Figure 4.3.4.) (b) Let f ( x) : = 1 I x2 for x -j. 0. The reader may show that lim ( J ix2 ) = 0 = X----+lim ( l lx2 ). (See Figure 4.3.3.) One x-__,.oo - oo way to do this is to show that if x 2: 1 then 0 :::; I I x2 :::; 1 I x. In view of part (a), this implies that lim ( l lx2 ) = 0. 0 X----> OC Just as it is convenient to be able to say that f( x) ---+ ±oo as x ---+ c for c E IR, it is convenient to have the corresponding notion as x ---+ ±oo. We will treat the case where X ---+ 00. 4.3.13 Definition Let A C:: IR and let f : A ---+ R Suppose that (a, oo) C:: A for some a E A. We say that f tends to oo [respectively, -oo] as x ---+ oo, and write lim f = oo X----> 00 [respectively, lim f = - oo] X --+ 00 if given any a E IR there exists K = K(a) > a such that for any x > K, then f(x) > a [respectively, f(x) < a]. As before there is a sequential criterion for this limit. 4.3.14 Theorem Let A E R let f : A ---+ R and suppose that (a, oo) C:: A for some a E R Then the following statements are equivalent: 122 CHAPTER 4 LIMITS (i) lim f = oo [respectively, lim f = -oo]. X----> 00 X----+ 00 (ii) For every sequence (x11 ) in (a, oo)such that lim(xn) = oo, then lim(f(xn)) = oo [respectively, lim(f(xn)) = -oo]. The next result is an analogue of Theorem 3.6.5. 4.3.15 Theorem Let A R let f, g : A R and suppose that (a, oo ) A for some -+ a E JR. Supposefurther that g(x) > Ofor all x > a and thatfor some L E JR, L -=f. 0, we have. f(x) lim - = L. x oc g ( X ) (i) If L > 0, then lim f = oo if and only if lim g = oo. x-___,. cx) X----+ CXJ lim00 f = - oo if and only if X----+ (ii) If L < 0, then X----+ lim00 g = oo. Proof. (i) Since L > 0, the hypothesis implies that there exists a 1 > a such that for Therefore we have G L) g(x) < f(x) < ( L) g(x) for all x > a 1 , from which the conclusion follows readily. The proof of (ii) is similar. Q.E.D. We leave it to the reader to formulate the analogous result as x -+ -oo. 4.3.16 Examples (a) lim x" = oo for n E.N. X---+ 00 Let g(x) := x" for x E (0, oo ). Given a E R let K : = sup{ 1 , a}. Then for all x > K, we have g(x) = x" 2: x > a. Since a E lR is arbitrary, it follows that lim g = oo. x oo (b) lim x" = oo for n E N, n even, and X----+lim - 00 x = - oo for n E N, n odd. " X----+ - 00 We will treat the case n odd, say n = 2k + 1 with k = 0, 1,.... Given a E R let K := inf{a, - 1 }. For any x < K, then since (x2 ) k 2: 1 , we have x" = (x2 l x :::; x < a. Since a E lR is arbitrary, it follows that X----+lim = - oo. - 00 (c) Let p : lR -+ lR be the polynomial function p (x) : = a11 X11 + an_ , xn - l + + a , x + ao. · · · Then lim p = oo if an > 0, and lim p = - oo if an < 0. X----+ 00 X --+ 00 Indeed, let g(x) := x" and apply Theorem 4.3. 1 5. Since it follows that x oo lim (p(x) jg(x)) = an. Since x oc lim g = oo, the assertion follows from Theorem 4. 3. 1 5. (d) Let p be the polynomial function in part (c). Then lim p = oo [respectively, -oo] if x->- 0. We leave the details to the reader. 0  4.3 SOME EXTENSIONS OF THE LIMIT CONCEPT 123 Exercises for Section 4.3 I. Prove Theorem 4.3.2. 2. Give an example of a function that has a right-hand limit but not a left-hand limit at a point. 3. Let f(x) := lxl - 1 1 2 for x =I 0. Show that lim0 + f(x) = lim f(x) = + oo. Let c E JR. and let f be defined for x E ( c, oo ) and f(x) > 0 for all x E ( c, oo ). Show that x-----+ x-o- 4. limf = oo if and only if lim 1 If = 0. X-----+ C X-----+ C 5. Evaluate the following limits, or show that they do not exist. X x I X- I (a) xlim -- (x =F I ) , lim __ (x =F I ), (b) x I+ X - 1 (c) x 0-1 lim (x + 2)lvfx (x > 0) , (d) xlim (x + 2)lvfx (x > 0) , - oo (e) lim ( J.X+I) Ix (x > - 1 ) , (f) X---- lim-+ 00 ( v'X+l) I x (x > 0) , x o vlx -5 vlx - x (g) (x > 0) , (h) xlim (x > O). }L Vx + 3 oo X+X 6. Prove Theorem 4.3. 1 1. 7. Suppose thatfand g have limits in JR. as x ----> oo and thatf(x) ::; g(x) for all x E ( a , oo ). Prove lim-+ OC.f ::; X--tCXJ that X---- lim g. 8. Let.f be defined on (0, oo ) to JR. Prove that x-->C G lim.f(x) = L if and only if x---- lim f( l lx) = L. Show that iff : ( a , oo ) ----> JR. is such that X---- lim-+ 00 xf(x) = L where L E JR., then Xlim -+ 0 + 9. ---700 f(x) = 0. 1 0. Prove Theorem 4.3. 1 4. I I. Suppose that xlimf(x) = L where L > 0, and that lim g(x) = oo. Show that limf(x)g(x) = oo. If L = 0, show by example that this conclusion may fail. -c x-----+ c x--c 1 2. Find functions f and g defined on (0, oo ) such that lim f = oo and x-oo lim g = oo, and X-----+ oo lim (f - g) = 0. Can you find such functions, with g(x) > 0 for all x E (0, oo ), such that x-CXJ lim flg = 0? X-----+ CXJ 1 3. Let f and g be defined on ( a, oo) and suppose lim f = L and lim g = Prove that lim f o g = L. oo..Y-->oo x--oo X----Jo OO