Continuity. Recall the following properties of limits. Theorem. Suppose that lim. f(x) =L and lim. lim. [f(x)g(x)] = LM, lim

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1 Recll the following properties of limits. Theorem. Suppose tht lim f() =L nd lim g() =M. Then lim [f() ± g()] = L + M, lim [f()g()] = LM, if M = 0, lim f() g() = L M. Furthermore, if f() g() for ll, then L M. Another useful result on limits is the squeeze, or sndwich, theorem: Theorem. Suppose f,g,h re functions such tht for ll (ecept possil t ) g() f() h(). Also ssume tht Then lim g() =limh() =L. lim f() =L. See slide for emple tht utilizes the squeeze theorem. Continuit Definition Afunctionf() iscontinuous t if lim f() =f(). There re 3 implictions of this def n: () the limit lim f() eists, Mth 30G - Prof. Kindred - Lecture 3 Pge

2 (2) f() isdefined, (3) nd the two ove quntities re equl. Emple Show tht f() =5 3iscontinuoust =. Lst clss, we showed tht lim (5 3) = 2 vi n δ proof. Furthermore, we know tht f() = 2, so it follows tht lim(5 3) = 2 = f(). Thus, f() iscontinuoust =. Question How cn function fil to e continuous t point? the limit t the point could fil to eist (see emples of how limits fil to eist from lst clss) the function m e undefined t tht point (f() undefined) lim f() = f() Emple Is f() = sin continuous t =0? f(!) π 5π 4π 3π 2π π 0 π 2π 3π 4π 5π 6π! No since f(0) is undefined. But we cn remove this discontinuit ecuse sin lim eists! 0 Clim: sin lim 0 =. We cn t use l Hopitl s rule, s we hve not proven it et. Mth 30G - Prof. Kindred - Lecture 3 Pge 2

3 Proof. Assume 0 << π.(thisisoksincewerelettinggo to 0.) 2 Consider the following regions nd their res: (, tn ) (cos, sin ) (cos, sin ) We hve sin cos 2 2π (π 2 ) () tn. 2 2 Then, multipling sin,weget cos sin cos = sin cos cos. We let 0 +,sothenthesqueezetheorem, sin lim. 0 + Thus, lim 0 + sin conclude tht Now, we define =. Notice tht sin F () = sin lim 0 =. sin if = 0 if =0. is n even function, so we An even function is function f such tht f( ) =f() for ll. The grph of n even function is smmetric with respect to the -is. Mth 30G - Prof. Kindred - Lecture 3 Pge 3

4 F () iscontinuoust =0. Furthermore,F () iscontinuoustll = 0sinceitisthertiooftwocontinuousfunctions. The function F () issometimesclled continuousetension off(). Definition We s f()iscontinuous on set S if f()iscontinuous t ever S. Importnt properties of continuous functions sums nd products of continuous functions re continuous (follows from limit properties) f() = is continuous nd f() =c for some constnt c is continuous (verif ourself s n eercise) using the two ove properties, we know tht n polnomil is continuous function Theorem (etreme vlue theorem). If f is continuous function on intervl [, ], then f hs n solute mimum t some vlue m [, ] nd f hs n solute minimum t some vlue min [, ]. f() f() f() m min f continuous ut onl on (, ) f not continuous on [, ] no m or min on this intervl no m or min on this intervl Mth 30G - Prof. Kindred - Lecture 3 Pge 4

5 Theorem (intermedite vlue theorem). If f is continuous function on [, ] nd 0 is n vlue etween f() nd f(), then there eists numer c [, ] such tht f(c) = 0. f() 0 The theorem seems firl ovious, ut it is not so es to prove (Mth 3 - Anlsis). c Note tht the theorem gurntees tht such numer c eists without stting how to find it. An importnt corollr follows... Corollr (eistence of roots). If f is continuous function on [, ] nd one of f() nd f() is positive nd other is negtive, then there eists vlue c [, ] such tht f(c) =0. Proof. Since f is continuous nd 0 is etween f() ndf(), it follows from the intermedite vlue theorem tht there eists t lest one vlue c (, ) suchthtf(c) =0. Mth 30G - Prof. Kindred - Lecture 3 Pge 5

6 Preservtion of continuit the theorem elow formlizes our previous sttement tht sums nd products of continuous functions re continuous. Theorem. Assume tht f() nd g() re continuous t point. Then the following functions re lso continuous t point : f() ± g() f()g(), f() g() if g() = 0. Mth 30G - Prof. Kindred - Lecture 3 Pge 6

7 Continuit Mth 30G, Clculus Professor Kindred Septemer 0, 202 Emple using squeeze/sndwich theorem We wnt to determine lim 0 2 sin We cnnot use the propert tht lim f ()g() = lim f () lim g() does not eist. since lim 0 sin. However, we do know tht sin for ll = sin 2 for ll = 0 nd since lim 0 2 = lim 0 2 = 0, the squeeze theorem, it must e tht lim 2 sin 0 = 0.