Lesson 10: Limits and Continuity

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1 Lesso 10: Limits ad Cotiuity SCIMS Academy 1

2 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals etc. as we will see later). To uderstad the cocept, cosider the fuctio f(x) = x 2. What value does the fuctio f approach, whe x approaches 2? It approaches 2 2 = 4. What does this mea? It meas that the differece betwee f(x) ad 4 ca be made as small as we please, if the differece betwee x ad 2 is made sufficietly small. We say that the limit of the fuctio x 2 as x approaches 2 is the value 4. I geeral, a fuctio f(x) has a limit L as x approaches a, if f(x) ca be made as close to L as we like, for all x sufficietly close to a. We write this as: lim f ( x) xa L Example: Let g x x x 2 ( ) lim g( x) lim(3x x 1) x1 x1 2 2 lim g( x) lim(3x x 1) x3 x3 Does it mea, that to evaluate the limit of f(x) as x approaches a, we just evaluate f(a)? Not always (the limit is f(a) oly for cotiuous fuctios, as we will see later). I fact, f(a) eed ot be defied for the limit to exist. SCIMS Academy 2

3 Limit of a fuctio (cotiued) Cosider the fuctio f(x) below (ote a fuctio ca have differet defiitios, i differet parts of its domai): f x x x 2 ( ) for 1, f ( x) 2 for x 1 2 lim f ( x) 1 1 but f (1) 2 x1 u( x) 1 for x 0 0 for x 0 ad To determie the limit of f(x) as x approaches a, we eed to kow the behavior of f(x) for x ear a. I the above example, eve if f(1) was udefied, the limit would still be 1. This is because the behavior of f(x) ear 1, depeds o the behavior of x 2 (the fuctio defiitio ear 1). I geeral, we do t defie limit for a poit x = a; istead we defie it for x approaches a. Cosider the uit step fuctio u(x) show below. What is its limit as x approaches 0? As you might have guessed, u(x) does ot have a limit as x approaches 0. Whe x approaches 0 from the right (values greater tha 0), u(x) has the limit 1 (called the right had limit); whe x approaches 0 from the left (values less tha 0), u(x) has the limit 0 (called the left had limit). Whe x approaches 0 from the left, u(x) is 0. Whe x approaches 0 from the right, u(x) is 1 SCIMS Academy 3

4 Right had ad Left had limits So by defiitio, f(x) has a right had limit L as x approaches a, if f(x) ca be made as close to L as we like, for all x greater tha but sufficietly close to a (x approaches a from the right alog the x axis). We write this as: lim f ( x) L xa Similarly, f(x) has a left had limit L as x approaches a, if f(x) ca be made as close to L as we like, for all x less tha but sufficietly close to a (x approaches a from the left alog the x axis). We write this as: lim f ( x) L xa For f(x) to have a limit as x approaches a, both the right had ad left had limit as x approaches a must exist ad they must be equal. The uit step fuctio u(x) does ot have a limit whe x approaches 0, sice the right had ad left had limit are differet. SCIMS Academy 4

5 Scearios whe the limit does ot exist For the greatest iteger fuctio, the right ad left had limits are ot equal (ad hece the limit does ot exist), whe x approaches ay iteger value. Whe x approaches 1 from the left, the fuctio has a value 0. Whe x approaches 1 from the right, the fuctio has a value 1. The fuctio f(x) = 1/x approaches ifiity as x approaches 0 from the right, ad approaches mius ifiity as x approaches 0 from the left. Ifiity is ot a umber, hece either the right had or left had limit exists whe x approaches 0. f(x) approaches ifiity, meas that f(x) ca be made larger tha ay value we choose; for all values of x greater tha but sufficietly close to 0. f( x) 1 x SCIMS Academy 5

6 Scearios whe the limit does ot exist (cotiued), ad usig ifiity i limits A less commo example of limit ot existig is the fuctio f(x) = si(1/x) as x approaches 0. To uderstad this example, you must kow how the si fuctio is defied for all values of x. The fuctio rapidly oscillates betwee 1 ad +1 (as we come close to 0), ad ever approaches ay specific value. So both the right had ad left had limit does ot exist whe x approaches 0. 1 f( x) si x Though ifiity is ot a umber, it is coveiet to use it i expressios as show below. 1 a) lim x0 x 1 b) lim 0 x x Example a) is what we have already see. Example b) says that 1/x ca be made as close to 0, as we please; by makig x sufficietly large. SCIMS Academy 6

7 Limit rules The limit rules ca be used to determie the limit, whe fuctios are combied via operatios of additio, multiplicatio etc. Limit rules Let lim f ( x) L ad lim g( x) M. The lim f ( x) g( x) L M xa lim f ( x) g( x) L M xa xa xa lim f ( x) g( x) LM xa f ( x) lim xa g ( x ) L M / / f x L lim ( ) r s r s xa (sum rule) (differece rule) (product rule) whe M 0 (quotiet rule) where r ad s are itegers (power rule) The above rules seem reasoable; for example, if f(x) approaches L ad g(x) approaches M, as x approaches a, we expect the sum f(x) + g(x) to approach L + M, as x approaches a. SCIMS Academy 7

8 Some limit examples lim x a xa lim k xa k where k is a costat If k is a costat ad is a positive iteger, the (applyig the product rule) lim kx lim k lim x x... x k lim x lim x... lim x xa xa xa times xa xa xa times If P( x) a x a x... a x a ka (applyig the sum rule, ad the above result) lim Px ( ) lim a x lim a x... lim a x lim a xa xa xa xa xa a a a a... aa a P( a) P( x) lim Px ( ) xa P( a) lim whe Q(a) 0 xaq ( x ) lim Q ( x ) Q ( a ) xa (polyomial fuctio), the If P(x) ad Q(x) are polyomial fuctios, the (usig quotiet rule ad above result) Note, ratio of two polyomial fuctios is called a ratioal fuctio. SCIMS Academy 8

9 Some limit examples (cotiued) If Q(a) = P(a) = 0, the (x a) is a commo factor, ad we ca cacel it out to evaluate the limit of the ratioal fuctio P(x) / Q(x) as x approaches a. If Q(a) = 0, ad P(a) 0, the the limit does ot exist as x approaches a. 2 x 4 2 Example: Evaluate lim. Both x 4 ad x 2 become zero at x 2 (called the 0/0 form). x2 x 2 x2 x2 But the limit ca be writte as lim x2 x 2 For x 2, we ca cacel x 2 ad write lim x 2 4. So the limit is 4. x2 I case, you are thikig why we added the coditio x 2, the cosider the defiitio of divisio. a/b by defiitio is ab 1, where b 1 is the multiplicative iverse of b. A umber x is the multiplicative iverse of y (ad vice versa) if xy = 1. All real umbers except 0, have a multiplicative iverse; hece divisio by zero is udefied. I the above example, divisio by (x 2) is udefied whe x = 2, so we add the coditio. From a limit viewpoit also, exclusio of x = 2 makes sese, because we are iterested i the behavior of the fuctio ear x = 2. SCIMS Academy 9

10 Some limit examples (cotiued) Example: Show that x a 1 l im a (where is a positive iteger). xa x a Solutio : We could divide x a by x a to get x a x a x x a x a... xa a (for x a) 1 Aother way is to treat the right side as a geometric series with first term x ad ratio a / x. 1 x 1 a / x x a We the have x x a x a... xa a 1 a / x x a x a lim lim x x a x a... xa a xa x a xa lim x lim x a... lim xa lim a a a... times a xa xa xa xa SCIMS Academy 10

11 Cotiuity of fuctios Let us ow cosider the closely related cocept of cotiuity of fuctios. A fuctio is cotiuous if the graph of the fuctio has o breaks: withi its domai, it is a cotiuous curve. A fuctio y = f(x) is cotiuous at a poit x = a i its domai if lim f ( x) f ( a) xa So by defiitio, the limit of a cotiuous fuctio as x approaches a is the same as f(a), a fact that we have used before to evaluate limits. Note that cotiuity (ulike limits) ca be defied for a poit. At a edpoit of the domai, the relevat oe sided limit is used i the defiitio. So at the left edpoit, it is the right had limit; ad at the right edpoit, it is the left had limit. This meas that if the fuctio domai is [a, b], the at x = a ad x = b, the fuctio is cotiuous if lim f ( x ) f ( a ) lim f ( x ) f ( b ) xa xb A fuctio is cotiuous o a iterval if it is cotiuous at every poit i the iterval. A cotiuous fuctio f(x) is cotiuous at every poit i its domai. This does ot imply that f(x) is cotiuous o ay iterval, because this may iclude poits outside the domai. However f(x) is cotiuous o ay iterval fully cotaied i its domai. If a fuctio f(x) is ot cotiuous at x = a, we say it is discotiuous at x = a, ad a is a poit of discotiuity of f. SCIMS Academy 11

12 Cotiuity of fuctios (cotiued) So if x = a is a poit of discotiuity for f(x), the oe of the followig holds: limit of f(x) as x approaches a does ot exist the limit exists, but is either ot equal to f(a), or f(a) is udefied. Whe f(a) is udefied, the poit x = a is ot part of the fuctio domai. But we ca still defie the limit of f(x) as x approaches a, if f(x) is defied ear a. Some examples are give below: Polyomial fuctio P(x) is cotiuous, as well as the ratioal fuctio P(x) / Q(x). Note poits where Q(x) = 0 are ot part of the fuctio domai. So f(x) = 1/x is a cotiuous fuctio. Absolute value fuctio x is cotiuous. The greatest iteger fuctio is discotiuous at all iteger values. The uit step fuctio u(x) is discotiuous at x = 0 (everywhere else it is cotiuous). Questio: Determie if the followig fuctio is cotiuous f(x) = 2x + 3 for x < 0 ad = ( 3 x) 2 for x 0. Solutio: Both ( 2x + 3) ad ( 3 x) 2 are polyomials, hece they are cotiuous. But ear x = 0, the differet defiitios of f(x) may ot approach the same value, hece there ca be a break. However this is ot so (therefore the fuctio is cotiuous), sice 2 lim 2 x 3 3 ad lim 3 x 3 ad f (0) 3. x0 x0 SCIMS Academy 12

13 Cotiuity rules If fuctios f(x) ad g(x) are cotiuous at x = a, the the followig combiatios are also cotiuous at x = a. f g x f x g x f g x f x g x a) b) f g c) f g ( x) f x g x d) (x) or (x) provided the deomiator is ot 0. g f r/ s r/ s e) f or g where r ad s are itegers (assumig the defiitio makes sese This readily follows from the limit rules, for example f g x f x g x lim ( ) lim ( ) ( ) by defiitio of fuctio sum xa xa = lim f ( x) lim g( x) by limit rule for a sum of two fuctios xa xa f ( a) g( a) sice f ad g are cotiuous at x a f g ( a) so the sum fuctio is cotiuous at x a Fuctio compositio: If g(x) is cotiuous at x = a, ad f(x) is cotiuous at g(a), the the composite f(g(x)) is cotiuous at x = a. ). SCIMS Academy 13

14 Properties of cotiuous fuctios Itermediate Value Theorem: If f(x) is cotiuous o a closed iterval [a, b], the it takes o every value betwee f(a) ad f(b). This meas that if y 0 is some value betwee f(a) ad f(b), the there exists a poit x = c i [a, b] such that f(c) = y 0. Extreme Value Theorem: If f(x) is cotiuous o a closed iterval [a, b], the f(x) has a absolute miimum m ad a absolute maximum M i [a, b]. This meas that: There exists x 1 ad x 2 i [a, b] such that f(x 1 ) = m ad f(x 2 ) = M, ad for all other x i [a, b], m f(x) M With the previous theorem, it implies that f(x) takes o every value betwee m ad M i [a, b]. f(x 2 ) = M = f(b) Graph of a arbitrary fuctio f(x) y 0 f(a) f(x 1 ) = m x = a x = c c x = x 1 x = b = x 2 SCIMS Academy 14

15 Explaatio o the two theorems The coditios stated i the Itermediate Value Theorem (IVT) ad Extreme Value Theorem (EVT) are ecessary, as we see below. Cosider f(x) = x i [0, 1) ad (x + 1) i [1, 2]. f(0) = 0 ad f(2) = 3. But f(x) does ot have ay value i [1, 2) because of the discotiuity at x = 1. We caot apply IVT to the iterval [0, 2], but we ca apply it to the iterval [1, 2]. Cosider f(x) = x defied o (0, 2). The iterval is ope, ad the fuctio does t have a absolute maximum or miimum value. If the domai is chaged to (0, 2], the f(x) has a absolute maximum of 2 at x = 2, but it has o absolute miimum. If the domai is chaged to [0, 2], the f(x) also has a miimum of 0 at x = 0. Thik about it, if the above is ot clear. For example, i the ope iterval (0, 2), we ca make x as close to 0 as we like, but we still have a ifiite umber of poits (like x/2, x/3,...) which are smaller tha x; hece there is o miimum. If f(x) = x i [0, 2) ad f(2) = 0 producig a discotiuity at x = 2, the f(x) has o absolute maximum. I summary, to apply EVT to a iterval, the iterval must be closed, ad f(x) must be cotiuous o that iterval. SCIMS Academy 15

16 Optioal: What follows is a more formal look at the limit cocept (eeds prior kowledge of iequalities, ivolvig absolute values). SCIMS Academy 16

17 Formal defiitio of limit We said that a fuctio f(x) has a limit L as x approaches a, if f(x) ca be made as close to L as we like, for all x sufficietly close to a. The formal defiitio give below defies closeess i a precise way. The fuctio f(x) has a limit L as x approaches a, if give ay positive ε (however small), we ca fid a positive δ, such that f(x) L < ε whe 0 < x a < δ. f(x) L < ε meas that f(x) lies i the ope iterval (L ε, L + ε). Let us call this iterval 1. By makig ε smaller, iterval 1 becomes smaller, ad f(x) stays closer to L. Similarly 0 < x a < δ meas that x lies i the ope iterval (a δ, a + δ) but is ever equal to a. Let us call this iterval 2. By makig δ smaller, iterval 2 becomes smaller, ad x stays closer to a. By choosig ε, we fix iterval 1 i which we wat f(x) to be preset. The defiitio says that whe the limit exists, we ca fid a iterval 2 aroud a, such that for all values of x i iterval 2; f(x) stays withi iterval 1. I other words, f(x) ca be kept as close to L as we like (less tha ε away from L), for all x sufficietly close to a (less tha δ away from a ). Note the defiitio does ot help us to fid L, but validates whether a give L is the limit (though the procedure is ot always straightforward). x1 Example: Cosider lim 2x Let us use the limit defiitio to prove that 5 is the limit. f ( x) L 2x x 1 x 1 / 2 Note the last iequality helps us to defie iterval 2. If δ = ε/2 or smaller, the f(x) 5 < ε whe 0 < x 1 < δ (which proves that 5 is the limit). SCIMS Academy 17

18 Formal defiitio (cotiued) Example: Prove lim x a for a 0. xa Solutio: We eed to fid such that x a whe 0 x a. x a x a. x a If we fid a positive costat C, such that x a C, ad if we choose C or smaller, we have x a x a x a x a C x a C x a Note the basic idea is to relate f(x) L to x a. Sice x > 0, we ca set C = a, ad δ = ε a or smaller (provig that a is the limit). The fuctio x is defied for all x 0, so at x = 0, oly the right had limit exists. A fuctio f(x) has a right had limit L as x approaches a, if give ay positive ε, we ca fid a positive δ, such that f(x) L < ε whe 0 < x a < δ (or equivaletly a < x < a + δ). Let us formally show that the right had limit of x is 0, as x approaches 0. x 0 < ε x < ε 2. Therefore δ ca be chose as ε 2 or smaller; which meas that the right had limit is 0. The above discussio also shows that x is a cotiuous fuctio. We ca come up with similar formal defiitios for other scearios: For example, we say that f(x) approaches ifiity as x approaches a, if for ay positive M (however large), we ca fid a positive δ, such that f(x) > M whe 0 < x a < δ. SCIMS Academy 18

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