1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point

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1 MA00 Capter 6 Calculus and Basic Linear Algebra I Limits, Continuity and Differentiability Te concept of its (p.7 p.9, p.4 p.49, p.55 p.56). Limits Consider te function determined by te formula f Note tat it is not defined at f as te form 0, wic is meaningless. We can, owever, still ask wat is appening to since at tis point f as approaces. More precisely, is f 0 approacing some specific number as approaces? To answer tis question, we do two tings: we calculate some values of y f as follows. f for near and sketc te grap of From te figure we see tat f In fact, approaces as approaces. In matematical symbols, we write (* Note tat as long as ). In general, to say tat f a l means tat te difference between f arbitrarily small by requiring tat be sufficiently close to but different from a. and l can be made

2 Eample can be calculated as follows:. Eample n n a n If n is a positive integer, we ave na. a a It is because, by long division, we ave n n a n n n n a a... a, a n wic tends to na as tends to a. Eample does not eist. It is because if if Terefore as approaces from te left, te values of approac te number. However, if approaces from te rigt, te values of approac te number. So it does not eist in tis case. From te eample we see tat te function tends to different values as approaces te number from different sides. In fact, given a function f, we can write f l (resp. f l ). a [read as te left (resp. rigt) and it of a f is l (resp. l ) ], if te values of (resp. l ) as approaces a from te left (resp. rigt). By using tis notation, we can write and. f tends to l Finally, we will use te symbol f l to represent te situation tat wen is increasing indefinitely, te values of te it is defined similarly. f l f tends to te value l,

3 Eample 4. It is because (* Note tat te values of tends to 0 as tends to positive infinity).. Teorems on its In te previous section, we ave introduced te notion of its. However in practical calculations, it is difficult to apply te definition directly. In tis section, a lot of tools will be introduced to elp us calculate te its. Te following teorems are typical eamples. Teorem A (Main Limit Teorem) Let n be a positive integer, k be a constant, and f and g be functions wic ave its at c. Ten. k k ; c. c ; c. kf k f ; c c 4. [ f g] f g ; c c c 5. [ f g] f g ; c c c 6. [ f g] f g; c c c 7. f f c, provided g c g g c c 0 ; n 8. f f c c ; 9. n f n f, provided c c n f 0 wen n is even. c Teorem B n n Let f an an m m aa0 and g bm bm b b0 be polynomials as sown. Ten we ave: (i) f f a n n a a a a n a a a n 0. a f f a (ii) If ga 0, ten. a g g a

4 Eample 5 t t 4 (i) Find. t t Te function wose it we are to find is te quotient of two polynomials. Te denominator t is not 0 wen t. Terefore, te it is te value of te quotient at t : t t4 4 t t 4 t 8 (ii) Find. t t 4 Te denominator is 0 at t and we cannot calculate te it by direct substitution. However, if we factor te numerator and denominator we find tat t 8 t t t4 t 4 tt t For t, t Terefore, for all values of t different from (te values tat really determine te it as t ), t 8 t t4 4 t t 4 t t 4 To calculate tis it, we divided te numerator and denominator by a common factor and evaluated te result at t. Teorem C (Sandwic Teorem or Squeeze Teorem) Suppose tat for all f t gt t t c in some interval about c and tat f t and f t l t ), ten approaces c (tat is, tc tc g t l. t c t approac te same it l as t Te following figure illustrates te idea beind te teorem. f(t) g(t) (t) Can you see te sandwic? c t-ais 4

5 Eample 6 Sow tat If we can sow ten we are done. sin 0 sin sin, 0 0 For te it sin, we consider te following diagram ( 0 ) 0 We see tat Tis implies Area OAP Area sector OAP Area OAT. sin tan. Furter simplification gives sin cos. As cos, by sandwic teorem, we ave 0 sin. 0 Tis is a rigt and it since we only consider 0. For were is positive, we see tat sin sin sin sin. Hence we ave sin sin. 0 0 Tus we can conclude tat sin. 0 5

6 Eample 7 sin Does eist? 0 sin 0 sin Observe tat. To te rigt of 0 and to te left of 0, te formulas are different so we sin 0 sin sin sin sin sin ave to consider,, separately. But, sin We conclude tat does not eist. 0 Eample 8 Find te it First, sin cos 0 sincos sin. Let t. Observe tat t tends to 0 as tends to 0. Terefore sin cos sin sint. 0 0 t0 t Eample 9 Evaluate 0 sin cos sin sin sin sin sin 0 cos 0 cos 0 cos sin sin sin sin cos cos 9. Eample 0 o sin o Evaluate, ( is te value of an angle measured in degree). 0 radians 80 It follows tat 6

7 o sin sin sin sin sin t t t t 0 80 Eample cos Evaluate. 0 sin cos cos sin sin sin cos 0 sin sin sin cos sin cos sin sin sin sint sin cos cos cos t t t cost Continuity of functions (p.64 p.7) In ordinary language, we use te word continuous to describe a process tat goes on witout abrupt canges. It is tis notion as it pertains to functions tat we now want to make precise. Consider te tree graps sown in te following figure. Only te tird grap eibits continuity at c. Here is te formal definition. y y y f f f c c c Definition (Continuity at a point) Let f be defined on an open interval containing c. We say tat f is continuous at c if f c f c By tis definition we require tree tings: f eists, () c () f c eists (tat is, c is in te domain of f ), and (). f f c c If any of tese tree fails, ten f is discontinuous at c. 7

8 Eample Te function 0 f 5 is undefined at 5. However, if we define 0 if 5 g 5 7 if 5 ten g will be continuous at 5. It is because 0 5 g 7 g Tus g becomes continuous at 5. Most of te functions tat we ere encountered are continuous. To name a few, we ave All polynomials are continuous at every real number c. Te absolute function is continuous at every real number c. sin and cos are continuous at every real number c. A rational function f g is continuous at every real number c wit gc 0 if f and Te following teorem will also elp us to identify tose continuous functions. g are bot continuous at c. Teorem A If f and g are continuous at c, ten so are kf, f g, f g, fg, f / g g c 0 and f n any real number and n is any positive integer., were k is Teorem B If g c l and if f is continuous at l, ten f g f g f l c c In particular, if g is continuous at c and f is continuous at at c. g c, ten te composite f g is continuous So far, we ave discussed continuity at a point. We can discuss continuity on an interval. We say f is continuous on an open interval if it is continuous at eac point of tat interval. It is continuous on te closed interval b, f f a ), and a, if it is continuous on ab, rigt continuous at a (tat is, left continuous at b (tat is, f f b b ). a 8

9 Continuous functions ave a lot of nice properties. One of tem is: Teorem C (Intermediate Value Teorem) If f is continuous on b a, and if l is a number between f c f a and f b, ten tere is a number c between a and b suc tat l. Tis means tat a continuous function must assume every value between any two of its values. Differentiability of functions (p.74 p.79, p.8 p.90) Consider te grap of te function y f as sown below: c, f c and c, f c Te points P and Q ave coordinates respectively. Te slope of te secant line PQ is given by slope of PQ f c f c As tends to 0, we ave te slope of te tangent at P is given by f c f c tan, 0 if suc a it eists. Tis leads to te concept of differentiability. A function f is said to be differentiable at c if te it f f c f c f c (or ) c c 0 eists. If so, it is called te derivative of f ' c. Again, we say tat f is f at c and is denoted by differentiable in an open interval I if it is differentiable at eac point of I. Eample If f, ten f f ' f So te function is differentiable at every real number ecept at 0. 9

10 Teorem A If f is differentiable at c, ten f is continuous at c. Te above teorem describes te relationsip between continuity and differentiability of a function. It sould be noted tat a continuous function may not be differentiable. Eample 4 Sow tat f, R is not differentiable at f '0 0 0 Tis it depends on te sign of. 0 0 Hence, (Note tat. 0 0 Te it does not eist. 0 f is continuous for all R ). Terefore, grap of y f. f is not differentiable at 0. Te geometric interpretation of tis can be seen from te For 0, te grap is te line y wit slope. For 0, te grap is te line y wit slope. At point O, were tese two sections of te grap meet, te grap as no well-defined slope. Tat is, te derivative is not defined at 0. 0

11 Eample 5 Let Sow tat Proof: f f is not differentiable at. for for. f f 0 0 f f f f does not eist. 0 f( ) is not differentiable at.