In this chapter, we define limits of functions and describe some of their properties.

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1 Chapter 2 Limits of Functions In this chapter, we define its of functions and describe some of their properties. 2.. Limits We begin with the ϵ-δ definition of the it of a function. Definition 2.. Let f : A R, where A R, and suppose that c R is an accumulation point of A. Then f() = L if for every ϵ > 0 there eists a δ > 0 such that 0 < c < δ and A implies that f() L < ϵ. We also denote its by the arrow notation f() L as c, and often leave it to be implicitly understood that A is restricted to the domain of f. Note that we eclude = c, so the function need not be defined at c for the it as c to eist. Also note that it follows directly from the definition that f() = L if and only if f() L = 0. Eample 2.2. Let A = [0, ) \ {9} and define f : A R by f() = 9 3. We claim that f() = 6. 9 To prove this, let ϵ > 0 be given. For A, we have from the difference of two squares that f() = + 3, and f() 6 = 3 9 = Thus, if δ = 3ϵ, then 9 < δ and A implies that f() 6 < ϵ.

2 2 2. Limits of Functions We can rephrase the ϵ-δ definition of its in terms of neighborhoods. Recall from Definition.6 that a set V R is a neighborhood of c R if V (c δ, c+δ) for some δ > 0, and (c δ, c + δ) is called a δ-neighborhood of c. A set U is a punctured (or deleted) neighborhood of c if U (c δ, c) (c, c + δ) for some δ > 0, and (c δ, c) (c, c + δ) is called a punctured (or deleted) δ-neighborhood of c. That is, a punctured neighborhood of c is a neighborhood of c with the point c itself removed. Definition 2.3. Let f : A R, where A R, and suppose that c R is an accumulation point of A. Then f() = L if and only if for every neighborhood V of L, there is a punctured neighborhood U of c such that A U implies that f() V. This is essentially a rewording of the ϵ-δ definition. If Definition 2. holds and V is a neighborhood of L, then V contains an ϵ-neighborhood of L, so there is a punctured δ-neighborhood U of c that maps into V, which verifies Definition 2.3. Conversely, if Definition 2.3 holds and ϵ > 0, let V = (L ϵ, L + ϵ) be an ϵ- neighborhood of L. Then there is a punctured neighborhood U of c that maps into V and U contains a punctured δ-neighborhood of c, which verifies Definition 2.. The net theorem gives an equivalent sequential characterization of the it. Theorem 2.4. Let f : A R, where A R, and suppose that c R is an accumulation point of A. Then f() = L if and only if f( n) = L. n for every sequence ( n ) in A with n c for all n N such that n = c. n Proof. First assume that the it eists. Suppose that ( n ) is any sequence in A with n c that converges to c, and let ϵ > 0 be given. From Definition 2., there eists δ > 0 such that f() L < ϵ whenever 0 < c < δ, and since n c there eists N N such that 0 < n c < δ for all n > N. It follows that f( n ) L < ϵ whenever n > N, so f( n ) L as n. To prove the converse, assume that the it does not eist. Then there is an ϵ 0 > 0 such that for every δ > 0 there is a point A with 0 < c < δ but f() L ϵ 0. Therefore, for every n N there is an n A such that 0 < n c < n, f( n) L ϵ 0. It follows that n c and n c, but f( n ) L, so the sequential condition does not hold. This proves the result. This theorem gives a way to show that a it of a function does not eist.

3 2.. Limits y 0 y Figure. A plot of the function y = sin(/), with the hyperbola y = / shown in red, and a detail near the origin. Corollary 2.5. Suppose that f : A R and c R is an accumulation point of A. Then f() does not eist if either of the following conditions holds: () There are sequences ( n ), (y n ) in A with n, y n c such that n = y n = c, but f( n) f(y n). n n n n (2) There is a sequence ( n ) in A with n c such that n n = c but the sequence (f( n )) does not converge. Eample 2.6. Define the sign function sgn : R R by if > 0, sgn = 0 if = 0, if < 0, Then the it sgn 0 doesn t eist. To prove this, note that (/n) is a non-zero sequence such that /n 0 and sgn(/n) as n, while ( /n) is a non-zero sequence such that /n 0 and sgn( /n) as n. Since the sequences of sgn-values have different its, Corollary 2.5 implies that the it does not eist. Eample 2.7. The it 0, corresponding to the function f : R \ {0} R given by f() = /, doesn t eist. For eample, consider the non-zero sequence ( n ) given by n = /n. Then /n 0 but the sequence of values (n) doesn t converge. Eample 2.8. The it ( ) sin, 0

4 4 2. Limits of Functions corresponding to the function f : R \ {0} R given by f() = sin(/), doesn t eist. (See Figure.) For eample, the non-zero sequences ( n ), (y n ) defined by n = 2πn, y n = 2πn + π/2 both converge to zero as n, but the its are different. f( n) = 0, n 2.2. Left, right, and infinite its f(y n) = n We can define other kinds of its in an obvious way. We list some of them here and give eamples, whose proofs are left as an eercise. All these definitions can be combined in various ways and have obvious equivalent sequential characterizations. Definition 2.9 (Right and left its). Let f : A R, where A R, and suppose that c R is an accumulation point of A. Then (right it) f() = L + if for every ϵ > 0 there eists a δ > 0 such that c < < c + δ and A implies that f() L < ϵ, and (left it) f() = L if for every ϵ > 0 there eists a δ > 0 such that c δ < < c and A implies that f() L < ϵ. Eample 2.0. For the sign function in Eample 2.6, we have sgn =, 0 + sgn =. 0 Net we introduce some convenient definitions for various kinds of its involving infinity. We emphasize that and are not real numbers (what is sin, for eample?) and all these definition have precise translations into statements that involve only real numbers. Definition 2. (Limits as ± ). Let f : A R, where A R. If A is not bounded from above, then f() = L if for every ϵ > 0 there eists an M R such that > M and A implies that f() L < ϵ. If A is not bounded from below, then f() = L if for every ϵ > 0 there eists an m R such that < m and A implies that f() L < ϵ.

5 2.2. Left, right, and infinite its 5 Sometimes we write + instead of to indicate that it denotes arbitrarily large, positive values, while denotes arbitrarily large, negative values. It follows from this definition that f() = f t 0 + ( ), f() = t f t 0 and it is often useful to convert one of these its into the other. Eample 2.2. We have + 2 =, + 2 =. ( ), t Definition 2.3 (Divergence to ± ). Let f : A R, where A R, and suppose that c R is an accumulation point of A. Then f() = if for every M R there eists a δ > 0 such that 0 < c < δ and A implies that f() > M, and f() = if for every m R there eists a δ > 0 such that 0 < c < δ and A implies that f() < m. The notation f() = ± is simply shorthand for the property stated in this definition; it does not mean that the it eists, and we say that f diverges to ±. Eample 2.4. We have Eample 2.5. We have 0 2 =, 0 + =, 2 = 0. 0 =. How would you define these statements precisely? Note that 0 ±, since / takes arbitrarily large positive (if > 0) and negative (if < 0) values in every two-sided neighborhood of 0. Eample 2.6. None of the its ( 0 + sin ), 0 sin ( ), 0 sin ( ) is or, since (/) sin(/) oscillates between arbitrarily large positive and negative values in every one-sided or two-sided neighborhood of 0.

6 6 2. Limits of Functions Eample 2.7. We have ( 3 ) ( ) =, 3 =. How would you define these statements precisely and prove them? 2.3. Properties of its The properties of its of functions follow immediately from the corresponding properties of sequences and the sequential characterization of the it in Theorem 2.4. We can also prove them directly from the ϵ-δ definition of the it, and we shall do so in a few cases below Uniqueness and boundedness. The following result might be taken forgranted, but it requires proof. Proposition 2.8. The it of a function is unique if it eists. Proof. Suppose that f : A R and c R is an accumulation point of A R. Assume that f() = L, f() = L 2 where L, L 2 R. Then for every ϵ > 0 there eist δ, δ 2 > 0 such that 0 < c < δ and A implies that f() L < ϵ/2, 0 < c < δ 2 and A implies that f() L 2 < ϵ/2. Let δ = min(δ, δ 2 ) > 0. Then, since c is an accumulation point of A, there eists A such that 0 < c < δ. It follows that L L 2 L f() + f() L 2 < ϵ. Since this holds for arbitrary ϵ > 0, we must have L = L 2. Note that in this proof we used the requirement in the definition of a it that c is an accumulation point of A. The it definition would be vacuous if it was applied to a non-accumulation point, and in that case every L R would be a it. Definition 2.9. A function f : A R is bounded on B A if there eists M 0 such that f() M for every B. A function is bounded if it is bounded on its domain. Equivalently, f is bounded on B if f(b) is a bounded subset of R. Eample The function f : (0, ] R defined by f() = / is unbounded, but it is bounded on any interval [δ, ] with 0 < δ <. The function g : R R defined by g() = 2 is unbounded, but is it bounded on any finite interval [a, b]. If a function has a it as c, it must be locally bounded at c, as stated in the net proposition.

7 2.3. Properties of its 7 Proposition 2.2. Suppose that f : A R and c is an accumulation point of A. If f() eists, then there is a punctured neighborhood U of c such that f is bounded on A U. Proof. Suppose that f() L as c. Taking ϵ = in the definition of the it, we get that there eists a δ > 0 such that 0 < c < δ and A implies that f() L <. Let U = (c δ, c) (c, c + δ), which is a punctured neighborhood of c. Then for A U, we have so f is bounded on A U. f() f() L + L < + L, Algebraic properties. Limits of functions respect algebraic operations. Theorem Suppose that f, g : A R, c is an accumulation point of A, and the its eist. Then f() = L, g() = M kf() = kl for every k R, [f() + g()] = L + M, [f()g()] = LM, f() g() = L M if M 0. Proof. We prove the results for sums and products from the definition of the it, and leave the remaining proofs as an eercise. All of the results also follow from the corresponding results for sequences. First, we consider the it of f + g. Given ϵ > 0, choose δ, δ 2 such that 0 < c < δ and A implies that f() L < ϵ/2, 0 < c < δ 2 and A implies that g() M < ϵ/2, and let δ = min(δ, δ 2 ) > 0. Then 0 < c < δ implies that f() + g() (L + M) f() L + g() M < ϵ, which proves that (f + g) = f + g. To prove the result for the it of the product, first note that from the local boundedness of functions with a it (Proposition 2.2) there eists δ 0 > 0 and K > 0 such that g() K for all A with 0 < c < δ 0. Choose δ, δ 2 > 0 such that 0 < c < δ and A implies that f() L < ϵ/(2k), 0 < c < δ 2 and A implies that g() M < ϵ/(2 L + ).

8 8 2. Limits of Functions Let δ = min(δ 0, δ, δ 2 ) > 0. Then for 0 < c < δ and A, f()g() LM = (f() L) g() + L (g() M) which proves that (fg) = f g. f() L g() + L g() M < ϵ 2K K + L ϵ 2 L + < ϵ, Order properties. As for its of sequences, its of functions preserve (non-strict) inequalities. Theorem Suppose that f, g : A R and c is an accumulation point of A. If f() g() for all A, and f(), g() eist, then f() g(). Proof. Let f() = L, g() = M. Suppose for contradiction that L > M, and let ϵ = (L M) > 0. 2 From the definition of the it, there eist δ, δ 2 > 0 such that f() L < ϵ if A and 0 < c < δ, g() M < ϵ if A and 0 < c < δ 2. Let δ = min(δ, δ 2 ). Since c is an accumulation point of A, there eists A such that 0 < a < δ, and it follows that f() g() = [f() L] + L M + [M g()] > L M 2ϵ > 0, which contradicts the assumption that f() g(). Finally, we state a useful sandwich or squeeze criterion for the eistence of a it. Theorem Suppose that f, g, h : A R and c is an accumulation point of A. If f() g() h() for all A and f() = h() = L, then the it of g() as c eists and g() = L.

9 2.3. Properties of its 9 We leave the proof as an eercise. We often use this result, without comment, in the following way: If 0 f() g() or f() g() and g() 0 as c, then f() 0 as c. It is essential for the bounding functions f, h in Theorem 2.24 to have the same it. Eample We have and but sin ( ) for all 0 ( ) =, 0 ( ) sin 0 =, 0 does not eist.