Econ 205 - Slides from Lecture 2Joel SobelAugust 24, 2010Econ 205 SobelLimits of FunctionsTake any function f ,f : X −→ Rfor X = (c, d) (so note that it is an open set) and take a ∈ X .Definitiony is called the limit of f from the right at a (right hand limit) if,for every ε > 0, there is a δ such that0 < x −a < δ =⇒| f (x) − y |< εthis is also denoted asy = limx→a+f (x)Note: x > a, so x is to the right of a.Econ 205 SobelLimits of FunctionsTake any function f ,f : X −→ Rfor X = (c, d) (so note that it is an open set) and take a ∈ X .Definitiony is called the limit of f from the left at a (left-hand limit) if, forevery ε > 0, there is a δ such that0 < a − x < δ =⇒| f (x) −y |< εthis is also denoted asy = limx→a−f (x)Note: x < a, so x is to the left of a.Econ 205 SobelLIMITDefinitiony is called the limit of f at a ify = limx→a−f (x)= limx→a+f (x)and we writey = limx→af (x)y is the limit of f at a if for every ε > 0, there is a δ > 0 such that0 <| x − a |< δ =⇒ | f (x) −y |< εNote: The definition of the limit of a function at the point a doesnot require the function to be defined at a.Econ 205 SobelExamplesConsider the functionf (x) =0, if x 6= 0,1, if x = 0.limx→0−f (x) = 0 = limx→0+f (x)⇒ limx→0f (x) = 0But the value of the function at the point 0 is not equal to thelimit. The example shows that it is possible for limx→af (x) toexist but for it to be different from f (a). Since you can define thelimit without knowing the value of f (a), this observation ismathematically trivial. It highlights a case that we wish to avoid,because we want a function’s value to be approximated by nearbyvalues of the function.Econ 205 SobelPropertiesTheoremLimits are unique. That is, if limx→af (x) = L andlimx→af (x) = L0, then L = L0.Proof.Assume that L 6= L0and argue to a contradiction. Letε =| L −L0| /2. Given this ε let δ∗> 0 have the property that| f (x) −L | and | f (x) −L0| are less than ε when 0 <| x − a |< δ∗.(This is possible by the definition of limits.) Since| f (x) −L | + | f (x) −L0|≥| L −L0| (1)it follows that ε ≥| L −L0|, which is not possible.Econ 205 SobelThe theorem simply states that the function cannot be close totwo different things at the same time. Inequality (inequality (1)) iscalled the triangle inequality. It says (in one dimension) that thedistance between two numbers is no larger that the distancebetween the first number and a third number plus the distancebetween the second number and the third number. In general, itsays that the shortest distance between two points in a straightline.Econ 205 SobelTheoremA limit exists if and only if for all ε > 0 there exists δ > 0 suchthat if0 <| x − a |< δ implies that | f (x) −f (a) |< εEcon 205 SobelTheoremIf f and g are functions defined on a set S, a ∈ (α, β) ⊂ S andlimx→af (x) = M and limx→ag(x) = N, then1. limx→a(f + g) (x) = M + N2. limx→a(fg) (x) = MN3. limx→afg(x) = M/N provided N 6= 0.For the first part, given ε > 0, let δ1> 0 be such that if0 <| x − a |< δ1then | f (x) −M |< ε/2 and δ2> 0 be such thatif 0 <| x − a |< δ2then | g(x) − N |< ε/2. This is possible by thedefinition of limit. If δ = min{δ1, δ2}, then 0 <| x − a |< δ implies| f (x)+g(x)−M −N |≤| f (x)−M | + | g (x)−N |< ε/2+ε/2 < ε.The first inequality follows from the triangle inequality while thesecond uses the definition of δ. This proves the first part of thetheorem.Econ 205 SobelFor the second part, set δ1so that if 0 <| x − a |< δ1then| f (x) −M |<√ε and so on.For the third part, note that when g(x) and N are not equal to zerof (x)g(x)−MN=g(x) (f (x) − M) + f (x) (N − g (x))g(x)N.So given ε > 0, find δ > 0 so small that if 0 <| x − a |< δ, then|f (x)||g(x)N|<2|M|N2, | f (x) −M |< Nε/2 and | g (x) −N |<N24|M|ε/2.This is possible provided that N 6= 0.Econ 205 SobelDefinitionWe say f : X → R is continuous at a iff (a) = limx→af (x)IIf the condition holds, then f is said to be continuous at thepoint a.IIf f is continuous at every point of X , then f is said to becontinuous on X .Ithe limit exists and the limit of the function is equal to thefunction of the limit. That is,limx→af (x) = f ( limx→ax).I state and prove a formal version of this below.IThe function has to be defined at a for the limit to exist.IIn order for a function to be continuous at a, it must bedefined “in a neighborhood” of a – that is, the function mustbe defined on an interval (α, β) with a ∈ (α, β). We extendthe definition to take into account “boundary points” in anatural way: We say that f defined on [a, b] is continuous at a(resp. b) if limx→a+f (x) = f (a) (resp. limx→b−f (x) = f (b)).Econ 205 SobelLemmaf : X −→ R is continuous at a if and only if for every ε > 0, thereis a δ > 0 such that0 <| a − x |< δ =⇒ | f (x) −f (a) |< εNote we could also write the conclusion as:| a − x |< δ =⇒ | f (x) −f (a) |< εEcon 205 SobelSequential ContinuityTheoremf : X −→ R is continuous at a if and only if for every sequence{xn} that converges to a, limn→∞f (xn) = f (a).Proof.Use the formulation of continuity in the previous lemma. Let ε > 0be given. Let δ > 0 correspond to that ε. Take a sequence {xn}that converges to a. We know that there is N such that if n > N,then | a −xn|< δ implies that | f (x) −f (a) |< ε. This proves thatcontinuity implies the sequential definition.Conversely, suppose that f is not continuous at a. It follows thatthere is ε > 0 such that for all δ > 0 there is x(δ) such that| x(δ) − a |< δ but | f (x(δ)) − f (a) |≥ ε. Let xn= x(1/n). Byconstruction, {xn} converges to a, but | f (xn) −f (a) |≥ ε so that{f (xn)} does not converge to f (a). We have shown that if f is notcontinuous at a then there is a sequence {xn} that converges to asuch that {f (xn)} doesn’t converge to f (a).Note that the condition in the theorem (limn→∞f (xn) = f …