Lecture 13: Limits of Functions

18.100A: Complete Lecture Notes Lecture 13: Limits of Functions Continuous Functions Remark 1. Continuous functions are those functions where tolerable changes to outputs accompany sufficiently small differences of inputs. Limits of Functions Definition 2 (Cluster Point) Let S ⊂ R. x ∈ R is a cluster point of S if ∀δ > 0, (x − δ, x + δ) ∩ S \ {x} 6= ∅. Let’s look at some examples. 1. S = {1/n | n ∈ N}. Here, 0 is a clusterpoint of S. 2. S = (0, 1). The set of cluster points of S is [0, 1]. 3. S = Q. The set of cluster points of S is R. 4. S = {0}. There are no cluster points of S. 5. S = Z. There are no cluster points of S. Theorem 3 Let S ⊂ R. Then, x is a cluster point of S if and only if there exists a sequence {xn } of elements in S \ {x} such that xn → x. Definition 4 (Function Convergence) Let S ⊂ R, let c be a cluster point of S, and f : S → R. We say that f (x) converges to L ∈ R at c if ∀ > 0 ∃δ > 0 such that if x ∈ S and 0 < |x − c| < δ, then |f (x) − L| < . Notation 5 Notationally, we may write f (x) → L as x → c, or limx→c f (x) = L. Theorem 6 Let c be a cluster point of S ⊂ R, and let f : S → R. If f (x) → L1 and f (x) → L2 as x → c, then L1 = L2 . 1 Proof : We will show ∀ > 0, |L1 − L2 | < . Let  > 0. Then, since f (x) → L1 and f (x) → L2 , ∃δ1 such that if x ∈ S and 0 < |x − c| < δ1 then |f (x) − L1 | < /2 and ∃δ2 > 0 such that if x ∈ S and 0 < |x − c| < δ2 , then |f (x) − L2 | < /2. Let δ = min{δ1 , δ2 } > 0. Then, since c is a cluster point of S, ∃x0 ∈ S such that 0 < |x0 − c| < δ =⇒ |L1 − L2 | = |L1 − f (x0 ) + f (x0 ) + L2 | ≤ |L1 − f (x0 )| + |f (x0 ) − L2 | < . Let’s see some examples of limits of functions. Example 7 Let f (x) = ax + b. Then, for all c ∈ R, limx→c f (x) = ac + b. Proof : Let  > 0. Choose δ =  1+|a| . Then, if x ∈ R and 0 < |x − c| < δ, then |f (x) − (ac + b)| = |a(x − c)| = |a||x − c| < |a|δ = |a|  < . 1 + |a|  Example 8 √ √ Let f (x) = x. Then, ∀c > 0, limx→c f (x) = c. √ Proof : Let  > 0. Choose δ =  c. Then, if x > 0 and 0 < |x − c| < δ, then |f (x) − √ √ √ c| = | x − c| √ √ √ √ ( x − c)( x + c) √ √ = x+ c |x − c| √ =√ x+ c |x − c| ≤ √ c δ < √ = . c  Example 9 1 x 6= 0 Let f (x) = . Then, limx→0 f (x) = 1. Notably, limx→0 f (x) 6= f (0)! 2 x = 0 2 Proof : Let  > 0 and choose δ = 1. Then, if 0 < |x − 0| < 1 then x 6= 0 =⇒ |f (x) − 1| = |1 − 1| = 0 < .  Question 10. How do limits of functions relate to limits of sequences? Theorem 11 Let S ⊂ R, c a cluster point of S, and let f : S → R. Then, the following are equivalent: 1. limx→c f (x) = L and 2. for every sequence {xn } in S \ {c} such that xn → c, we have f (xn ) = L. Proof : (1. =⇒ 2.): Suppose limx→c f (x) = L. Let {xn } be a sequence in S \ {c} such that xn → c. We want to show that f (xn ) → L. Let  > 0. Given limx→c f (xn ) = L, ∃δ > 0 such that if x ∈ S and 0 < |x − c| < δ then |f (x) − L| < . Since xn → c, ∃M0 ∈ N such that ∀n ≥ M0 , 0 < |xn − c| < δ. Choose M = M0 . Then, ∀n ≥ M , if 0 < |xn − c| < δ then |f (xn ) − L| < . Thus, f (xn ) → L. (2. ⇐= 1.): Suppose 2. holds, and assume for the sake of contradiction that 1) is false. Then, ∃0 > 0 such that ∀δ > 0, ∃x ∈ S such that 0 < |x − c| < δ and |f (x) − L| ≥ 0 . Then, ∀n ∈ N, ∃xn ∈ S such that 0 < |xn − c| < 1 n and |f (xn ) − L| ≥ 0 . By the Squeeze Theorem applied to 0 < |xn − c| < 1 , n xn → c. Then, by 2., 0 = lim |f (xn ) − L| ≥ 0 n→∞ which is a contradiction. 3