Lecture notes in basic Mathematics pdf

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Veer Surendra Sai University of Technology, Burla Department of Mathematics Lecture notes in Mathematics-I (Dr. S. Mohapatra) 1 Limit of function: Letf beareal-valuedfunctionde nedforallpointsinaneighbourhoodN ofapointcexcept possibly at the point c itself. Recall that any open set containing the element a is called the neighbourhoodofa. InparticularN(a;) = (a;a+); 0iscalledthe-neighbourhood  of a and N (a;) = N(a;)fag = (a;a)(a;a+) is called the deleted neighbourhood of a. We are assuming that f is de ned in the deleted neighbourhood of a in the following de nition of limit. De nition 1 The function f : X R is said to have a limit at x = a (a may or may not belongs to X) if given " 0, there exists , depending upon a and ", and there exists L2R such that 0 jxaj )jf(x)Lj "; that is, x2 (a;a)(a;a+)) f(x)2 (L";L+"); that is,  x2 N (a;)) f(x)2 N(L;"): When this happens, we say that the limit L of f exists at x = a and write it as limf(x) = L; or;f(x) L as x a: xa This de nition is called the " de nition of limit. One may observe that f need not be de ned as a; even it is de ned, it is not necessary that f(a) be equal to L. De nition 2 The functionf is said to tend to+1 asx tends toc (or in symbols, limf(x) = xc +1) if for each G 0 (however large), there exists a  0 such that f(x) G; whenever jxcj : Similarly,thefunctionf issaidtotendto1asxtendstoc(orinsymbols, limf(x) =1) xc if for each G 0 (however large), there exists a  0 such that f(x) G; whenever jxcj : De nition 3 The function f is said to tend to a limit l as x tends to 1 (or in symbols, lim f(x) = l) if for each " 0, there exists a k 0 such that x1 jf(x)lj "; whenever x k: De nition 4 The function f is said to tend to +1 as x tends to 1 (or in symbols, lim f(x) =1) if for each G 0 (however large), there exists a k 0 such that x1 f(x) G; whenever x k:1.1 Left hand and right hand limits While de ning the limit of a function f(x) as x tends to c, we consider the values of f(x) when x is very close to c. he values of x may be greater or less than c. If we restrict x to values less than c, then we say that x tends to c from below or from the left and write it symbolically as x c0 or simply x c. he limt of f(x) with this restriction on x, is called the left hand limit. Similarly, if x takes only the values greater than c, then x is said to tend to c from above or from the right, and is denoted symbolically as x c+0 or x c+. The limit of f(x) with this restriction on x, is called the right hand limit.  De nition 5 A function f is said to tend to a limit l as x tends to c from the left, if for each " 0, there exists a  0 such that jf(x)lj "; whenever c x c: In symbols, we then write lim f(x) = l: xc De nition 6 A function f is said to tend to a limit l as x tends to c from the right, if for each " 0, there exists a  0 such that jf(x)lj "; whenever c x c+: In symbols, we then write lim f(x) = l: xc+ Note: We say limf(x) exists if and only if both the limits (the left hand and the right hand) xc exists and are equal. One-sided in nite limit may also be de ned in the same way as above. Example 1 Let 8 2 2 x a ; x = ̸ a; f(x) = xa : b; x = a: Then show that limf(x) = 2a, by using " de nition. xa Sol. This function de ned on R. Take " 0. The condition that 0 jxaj , implies x = ̸ a. So we may write 2 2 x a (xa)(x+a) = = x+a: xa xa Now jf(x)2aj =j(x+a)2aj =jxaj Since we need jf(x)2aj " whenever jxaj , clearly we can choose any  such that 0  ". Hence it follows from the de nition that 2 2 x a lim = 2a: xa xa 2 2 x a Note that 2a = lim ̸= f(a) = b, unless b = 2a. xa xa 23 3 x a 2 Example 2 Show that lim = 3a , by using " de nition. xa xa 3 3 x a 2 2 Sol. Let " 0. Take x̸= a. Then f(x) = = x +xa+a and hence xa 3 3 x a 2 2 2 3a = j(x a )+a(xa)j xa  jxajjx+2aj  jxaj(jxaj+3jaj) As we need  0 such thatjxaj , choosing rst  1, the right hand side of the above inequality is less than or equal to (1+3jaj). This gives an idea as to what  can be choosen for a given " 0. ( ) " " Choose  = min 1; so that  . 1+3jaj 1+3jaj Now 2 jf(x)3a j (1+3jaj) ": This proves the result. We observe that the choice of  depends not only on ", but also on a. Although the natural domain of f isRfag. p Example 3 Show that lim 4x+1 = 3, by using " de nition. x2 Sol. Let " 0 be given. Now p p p ( 4x+13)( 4x+1+3) p j 4x+13j = 4x+1+3 4jx2j = p 4x+1+3 1 1 Choose  1. Sojx2j  1 implies 1 x 3. Hence p p 4x+1+3 5+3 So p 4 p j 4x+1+3j : 5+3 p 4 ( 5+3)" p In order that ", we choose  . 4 5+3 p p ( 5+3)" ( 5+3)" Now nally we choose  = min(1; ) so that  . 4 4 p 4 p Hencej 4x+13j ". 5+3 This proves the result. 1 Example 4 Evaluate lim : 1=x x0+ 1+e 1=x 1=x Sol. As x 0+, we feel that 1=x increases inde nitely, e increases inde nitely. e tends to 1; thus the required limit may be 1. We have to show that for a given " 0;9 a  0 such that 1 1 "; whenever 0 x : 1=x 1+e 3 1=x 1 e 1 1 1 1 1=x Now 1 = = ", when 1+e or log( 1) " x " 1=x 1=x 1=x 1+e 1+e 1+e 1 ) 0 x ; for 0 " 1: log(1="1) 1 1 Thus choosing  = , we see that if 0 " 1, 1 "; when 0 x . 1=x log(1="1) 1+e 1 1 1=x Again when " 1; 1 ") e 1, which is true for all values of x, so that 1=x " 1+e for any  0 would work. 1 Thus for any " 0 we are able to nd a  0 such that 1 ", when 0 x . 1=x 1+e 1 ) lim = 1. 1=x x0+ 1+e 1 Example 5 Prove that limxsin = 0. x0 x 1 1 Sol. Now xsin =jxj sin jxj x x 1 Thus choosing a  = ", we see that xsin ", when 0 jxj . x 1 ) limxsin = 0: x0 x 1 Example 6 Show that lim =1. 4 x3 (x3) Sol. Let G be any positive number, however large. 1 1 4 1 1 Now G, or G, when (x3) or when 0 jx3j . 1=4 4 4 G G (x3) (x3) 1 Choosing  = , we get the required result. 1=4 G 1=(x1) Example 7 Show that lim2 does not exist. x1 Sol. We rst consider the left hand limit. Let " 0 be given. Choosen a positive integer m m such that 1=2 ". 1 1 1 Take  = and let x satisfy 1 x 1. Now (x1) 0, and so 0. m x1  1=(x1) 1=(x1) 1= m 1=(x1) Thusj2 0j = 2 2 2 " and hence lim 2 = 0. x1 Next, consider x to be on the right of 1. 1 Let  0 be arbitrary and choose a positive integer m such that . Then if 0 m 0 1 1 1 n 1=(x1) 1+ 1 n n m ; 1+ 2 (1;1+) and 2 = 2 , which is unbounded. Therefore lim2 does 0 n x1 not exist. Example 8 Find the right and the left hand limits of a function de ned as follows 8 jx4j ; x = ̸ 4; f(x) = x4 : 0; x = 4: 4jx4j x4 Sol. When x 4;jx4j = x4. ) lim f(x) = lim = lim = lim 1 = 1: x4+ x4+ x4+ x4+ x4 x4 (x4) Again, when x 4;jx4j =(x4). ) lim f(x) = lim = lim (1) =1: x4 x4 x4 x4 so that lim f(x)̸= lim f(x): x4+ x4 Hence limf(x) does not exist. x4 Example 9 If limf(x) exists, prove that it must be unique. xa Sol. Let if possible, f(x) tend to limits l and l . Hence for any " 0 it is possible to choose 1 2 a  0 such that jf(x)l j "=2, when 0 jxaj . 1 jf(x)l j "=2, when 0 jxaj . 2 Now jl l j =jl f(x)+f(x)l jjl f(x)j+jf(x)l j ", when 0 jxaj . 1 2 1 2 1 2 i.e.,jl l j is less than any positive number " (however small) and so must be equal to zero. 1 2 Thus l = l . 1 2 Theorem 1 (without proof) If f and g are two real valued functions de ned on some neigh- bourhood of c such that limf(x) = l and limg(x) = m then xc xc (i) Let 2R. We have lim f(x) = l. xc (ii) lim(fg)x = limf(x)limg(x) = lm. xc xc xc (iii) lim(fg)x = limf(x):limg(x) = lm: xc xc xc (iv) lim(f=g)x = limf(x)=limg(x) = l=m; provided m̸= 0. xc xc xc p (x+2)(3x1) 4+x2 sinx Example 10 Evaluate (i) lim , (ii) lim , (iii) lim p . 2 x1 x0 x0+ x +3x2 x x lim (x+2): lim (3x1) (x+2)(3x1) 1:(4) x1 x1 Sol. (i) lim = = = 1. 2 2 x1 x +3x2 4 lim x +3x2 x1 p p p 4+x2 4+x2 4+x+2 1 1 (ii) lim = lim :p = limp = . x0 x0 x0 x x 4 4+x+2 4+x+2 ( )( ) p sinx sinx p (iii) lim = lim : lim x = 1:0 = 0. x0+ x0+ x0+ x x 2 x 1 Example 11 Evaluate lim . x1 x1 Sol. Let us evaluate the left hand and right hand limits. When x 1, put x = 1h; h 0. 2 2 (x 1 (1h) 1 h(2h) lim = lim = lim = lim (2h) = 2. x1 x1 h0+ h h0+ h h0+ Again when x 1+, put x = 1+h; h 0. 2 2 (x 1 (1+h) 1 lim = lim = lim (2+h) = 2. x1+ x1 h0+ h h0+ So that both, the left hand and the right hand, limits exist and are equal. Hence limit of the given function exists and equals to 2. 51=x e Example 12 Evaluate lim . 1=x x0 e +1 1=x 1=x Sol. Now when x 0+;1=x1;e 0 and x 0;1=x1;e 0. 1=x e 1 ) lim = lim = 1. 1=x 1=x x0+ x0+ e +1 e +1 1=x e 0 and lim = = 0. 1=x x0 e +1 1 1=x e so that the left hand limit not equal to the right hand limit. Hence lim does not 1=x x0 e +1 exist. x Example 13 Find lime sgn(x+x), where the signum function is de ned as x0 8 1; if x 0; sgn(x) = 0; if x = 0; and x means the greatest integer  x. : 1; if x 0; x 0h h Sol. Now lim e sgn(x+x) = lim e sgn(0h+0h) = lim (e ) =1; x0 h0+ h0+ x 0+h h lim e sgn(x+x) = lim e sgn(0+h+0+h) = lim e = 1. x0+ h0+ h0+ x ) lime sgn(x+x) does not exist. x0 Assignment 1 1. Prove the following limits by using " de nition: 2 x 4 (i) lim = 2, 2 x2 x 2x p (ii) lim x+3 = 3, x6 n n (iii) lim x = a , xa 2 2 (iv) limf(x) = a , where f :RR be given by f(x) = x , xa 1 (v) limf(x) = 1=a, where f : (0;1)R be given by f(x) = x . xa 2. Evaluate the following limits (i)-(vi), if they exist: ( ) 3x+jxj 1 1 2 (i) lim , (ii) lim , x0 x1 7x5jxj x1 x+3 3x+5 x x 12cosx+cos2x e e (iii) lim , (iv) lim , 2 x x x0 x1 x e +e tanxx 1+cosx (v) lim , (vi) lim , 2 x0 x1 x(1cosx) tan x 1=x xe (vii) Show that lim = 0, 1=x x0 1+e 1=x e 1 (viii) Show that lim does not exist, 1=x x0 e +1 1=x 1=x e e (ix) Show that lim does not exist, 1=x 1=x x0 e +e (x) If limf(x) = l then show that limjf(x)j =jlj. xc xc 1.2 Limit of a function by sequential approach De nition 7 Let J  R be an interval. Let a 2 J. Let f : Jnfag R be given. Then limf(x) = l iff for every sequence fx g with x 2 Jnfag with the property that x a, we n n n xa have f(x ) l. n 6Theorem 2 A function f tends to nite limit as x tends to c if and only if for every " 0 9 a neighbourhood N(c) of c such thatjf(x )f(x )j " for all x ;x 2 N(c);x ;x ̸= c. m n m n m n Similarly, a function f tends to a nite limit as x tends to 1 if and only if for every " 0, there exists G 0 such thatjf(x )f(x )j ", for all x ;x G. m n m n 1 1 Example 14 Show that lim sin does not exist. x0 x x 1 1 Sol. Let f(x) = sin . The function f is de ned for every non-zero real number. x x (4n+1) 2  Now for each natural number n, let x = , and so f(x ) = sin(2n + ) = n n (4n+1) 2 2 (4n+1) 1 as n1. 2 2 ) lim f(x ) =1, whenfx g =f g converges to zero. n n (4n+1) n1 Again, by taking x = 1=n, we see that f(x ) = n:0 = 0 for every natural number n, and n n so lim f(x )̸=1, whenfx g =f1=ng converges to zero. n n n1 Therefore, limf(x) does not exist. x0 2 Example 15 Find lim x sgn(cosx). x1 2 Sol. Here, f(x) = x sgn(cosx). Let x =2n, sofx g1, as1. n n 2 2 2 Now f(x ) = (2n) sgn(cos(2n)) = 4n  , and so lim f(x ) = 1, when fx g = n n n n1 f2ng1. 2 ) lim x sgn(cosx) =1. x1 2 Again, taking x =(2n+1), we see that f(x ) = (2n+1) sgn(cos((2n+1))) = n n 2 2 2 (2n+1)  and so lim x sgn(cosx) =1. x1 2 Hence lim x sgn(cosx) does not exist. x1 Assignment 2 (i) The function f :Rnf0g 1;1 de ned by f(x) = sin(1=x) does not exist at x = 0 by using sequential approach. 3 2 (ii) Use sequential limit form to obtain the limit of f(x) = x +x 5 at x = a. Lemma 1 (Sandwich Theorem) Let J R. Let f;g;h be de ned on Jnfag. Assume that (i) f(x) h(x) g(x); for x2 J;x̸= a. (ii) limf(x) = l = limg(x). xa xa Then limh(x) = l. xa 2 Continuity Let f be a real-valued function de ned on an interval J  R. We shall now consider the behaviour of f at points on J. De nition 8 (" de nition of continuity) Let f : J R be given and a2 J. We say that f is continuous at a if for any given " 0, there exists  0 such that x2 J andjxaj  )jf(x)f(a)j ". De nition 9 A function f(x) is said to be continuous at a point c 2 J, if limf(x) exists xc and the limit equals to the value of the function at x = c (i:e:; limf(x) = f(c)). xc 7A function f is said to be continuous in an interval J, if it continuous at every point of the interval. Afunctionissaidtobediscontinuousatapointx = cofitsdomain, ifitisnotcontinuous at x = c. The point x = c is called a point of discontinuity of the function. Types of discontinuities: (i) A function f is said to be have a removable discontinuity at x = c, if limf(x) exists xc but is not equal to the value f(c) (which may or may not exist) of the function. Such a discontinuity can be removed by assigning a suitable value to the function at x = c. (ii) A function f is said to have a discontinuity of the rst kind at x = c, if lim f(x) and xc lim f(x) both exist but are not equal. xc+ (iii) A function f is said to have a discontinuity of the rst kind from the left at x = c, if lim f(x) exists but is not equal to f(c). xc (iv) A function f is said to have a discontinuity of the rst kind from the right at x = c, if lim f(x) exists but is not equal to f(c). xc+ (v) A function f is said to have a discontinuity of the second kind at x = c, if neither lim f(x) nor lim f(x) exists. xc xc+ (vi) A function f is said to have a discontinuity of the second kind from the left at x = c, if lim f(x) does not exist. xc (vii) A function f is said to have a discontinuity of the second kind from the right at x = c, if lim f(x) does not exist. xc+ Theorem 3 (without proof) Let f;g : J R be continuous at a point a 2 J. Let 2 R. Then the functions f;jfj;f+g;fg;fg are also continuous at x = a and if g(a) = ̸ 0, then f=g is also continuous at x = a. De nition 10 Let f;g : J R be a real-valued function and a 2 J. Assume that f is continuous at a and g is continuous at f(a). Then the composition function (gof) is also continuous at a. De nition 11 Let J R. Let f : J R be a real-valued function and a2 J. We say that f is continuous at a if for every sequence fx g in J with x a, we have f(x ) f(a). n n n Example 16 Discuss the continuity of the following functions: (i) f(x) = 1=x, (ii) f(x) = sin(1=x). Sol. (i) The function f(x) = 1=x has the natural domainRnf0g. Here lim f(x) =1 and lim f(x) =1: x0+ x0+ Thus f(x) has a discontinuity of the second kind at x = 0. (ii) The function f(x) = sin(1=x) has the natural domain Rnf0g. It has been observed from Q. 1 in Assignment- 2, that it does not have a limit as x 0. In fact lim f(x) and lim f(x) do not exist as f(x) oscillates between 1 and1, as x 0. x0+ x0 Thus f(x) has a discontinuity of the second kind at x = 0. 8Example 17 Let xsin(1=x); if x̸= 0; f(x) = 0; if x = 0: Show that f is continuous at x = 0. Sol. Sincejxsin(1=x)jjxj, it follows from the Lemma 1 (Sandwich Theorem) that limxsin(1=x) = 0: x0 Thus limf(x) = f(0). x0 Hence the function f is continuous at x = 0. Example 18 Discuss the continuity of 8 1+x; 1 x 0; f(x) = 1+x+sinx; 0 x =2; : 3; x =2: Sol. Since f(0) = 1 = lim f(x) = lim (1+x) = 1 = lim f(x) = lim (1+x+sinx) = 1. x0 x0 x0+ x0+ It follows that f is continuous at x = 0. Since x is right continuous but not left continuous at x = 1, so also is f. Again f(=2) = 3 = lim f(x), x=2+ lim f(x) = lim (1+x+sinx) = 1+ lim x+ lim sinx = 1+1+1 = 3. x=2 x=2 x=2 x=2 Hence f is also continuous at =2. Thus f is continuous at every point ofR except at x = 1. Example 19 Discuss the kind of discontinuity (if any) of the function de ned as follows: xjxj ; x̸= 0; f(x) = x 2; x = 0: x+x Sol. Now lim f(x) = lim = 2, x0 x0 x xx lim f(x) = lim = 0, x0+ x0+ x and f(0) = 2. Thus the function has discontinuity of the rst kind from the right at x = 0. 8 1=x xe ; x = ̸ 0; 1=x Example 20 Show that the function de ned by f(x) = is continuous 1+e : 0; x = 0; at x = 0. 1=x xe Sol. Now lim f(x) = lim = 0, 1=x x0 x0 1+e x lim f(x) = lim = 0, 1=x x0+ x0+ e +1 and f(0) = 0. ) limf(x) = 0 = f(0). x0 Thus the function is continuous at x = 0. 9Example 21 Discuss the continuity of sin2x ; x̸= 0; f(x) = x 1; x = 0; at x = 0. sin2x Sol. Now limf(x) = lim :2 = 2, so that limf(x) = ̸ f(0). x0 x0 x0 2x Hence the limit exists, but is not equal to the value of the function at the origin. Thus the function has a removable discontinuity at x = 0. 1; x2Q; Example 22 Let f :RR be given by f(x) = Then f is not contin- 0; x2RQ: uous at any point of R. This is known as Dirichlet's function. Sol. Let a2Q so that f(a) = 1. Since in any interval there lie an in nite number of rational and irrational numbers, therefore 1 for each positive integer n, we can choose an irrational number a such that ja aj . n n n Thus the sequencefa g converges to a. n But f(a ) = 0 for all n and f(a) = 1, so that the sequence f(a ) does not converge to n n f(a) (i.e., lim f(a )̸= f(a)). Thus we conclude that f is not continuous at all a2Q. n x1 Next, let b2RQ. For each positive integer n we can choose a rational number b such n 1 thatjb bj . Thus the sequencefb g converges to b. n n n But f(b ) = 1 for all n and f(b) = 0, so that the sequence f(b ) does not converge to n n f(b) (i.e., lim f(b )̸= f(b)). Thus we conclude that f is not continuous at all b2RQ. n x1 ) f is continuous nowhere onR. Assignment 3 1. Discuss the continuity and classify the discontinuities, if any, of the following functions; 8 sin(xc) xsin(1=x); x = ̸ 0; ; x = ̸ c; (i) f(x) = (ii) f(x) = xc 0; x = 0; : 0; x = c; 2 1 1 x ; x 0; (iii) f(x) = (iv) f(x) = cosec , x; x 0; xa xa 8 1=x e 1=x (1+x) ; x = ̸ 0; ; x̸= 0; 1=x (v) f(x) = (vi) f(x) = 1e 1; x = 0; : 1; x = 0; 8 1=x 1=x e e x; x2Q; ; x = ̸ 0; 1=x 1=x (vii) f(x) = (viii) f(x) = e +e x; x2RQ: : 1; x = 0; 8 0; x = 0; 8 1=x (1=2)x; 0 x 1=2; e 1 ; x̸= 0; (ix) f(x) = 1=x (x) f(x) = 1=2; x = 1=2; e +1 : 0; x = 0; 3=2x; 1=2 x 1; : 1; x = 1; 8 3 x 8 2x; x 1; ; x = ̸ 2; 2 (xi) f(x) = (xii) f(x) = x 4 2 x ; x 1: : 3; x = 2; 103 Differentiability of functions De nition 12 Let J be an interval and c 2 J. Let f : J R. Then f is said to be differentiable at c, if there exists a real number such that f(x)f(c) lim = : (1) xc xc It is sometimes useful to use the variable h for the increment xc and reformulate (1) as follows: f(c+h)f(c) lim = : (2) h0 h ′ The limit value is called the derivative of the function f at x = c and is denoted by f (c). It should be noted that before examining the differentiability of f at c, it is necessary to ensure that f is de ned in a neighbourhood of c. Let J be an open interval and let f : J R be differentiable at every point of J, then ′ we can de ne a new function f : J R de ned by f(y)f(x) ′ f (x) = lim : yx yx ′ Thus we get an operator D (say), which takes f to f whenever f is differentiable. Thus ′ Df = f . 3.1 Left and right derivative De nition 13 Let f : J R; c 2 J. The function f is said have a left derivative at the point x = c if there exists m2R such that f(x)f(c) lim = m; xc0 xc ′ ′ ′ and m is said to be left derivative of f at c and we denote it as f (c0);f (c), Lf (c) or D f(c). Similarly, the function f is said have a right derivative at the point x = c if there exists m 2R such that 0 f(x)f(c) lim = m ; 0 xc+0 xc ′ ′ ′ and m is said to be right derivative of f at c and we denote it as f (c+0);f (c+);Rf (c) or 0 D f(c). + NOTE: If a function is differentiable at x = c, then (i) D f(c) should exist. (ii) D f(c) should exist. + (iii) D f(c) = D f(c) = Df(c). + Thus a function f is not differentiable if any one of the above requirements is not met. In view of the de nition of limit, the differentiability condition in (1) can be de ned using " as follows: 11De nition 14 We say that f is differentiable at c if there exists 2 R such that for any given " 0, there exists a  0 such that x2 J and 0 jxcj )jf(x)f(c) (xc)j "jxcj: (3) We say that f is differentiable on J if it is differentiable at each c2 J. Example 23 Let f : J R be a constant function, say, C. Then f is differentiable at ′ c2 J with f (c) = 0 by using " de nition. Sol. Consider the expression f(x)f(c) CC = = 0: xc xc ′ ′ This suggests that = f (c) = 0. Now we will prove that f (c) = 0 by using " de nition. Let " 0 be given. Now let us try to estimate the error term: jf(x)f(c) (xc)j =jCC (xc)j =j jj(xc)j = 0: This suggests that we can choose any  0 for any " 0. Let " 0 be given. Let  0 be arbitrary. We can estimate the error term: jf(x)f(c) (xc)j = 0 "jxcj: ′ Thus if f is a constant function then it is differentiable in R with f (c) = 0 for all c2 J R. Example 24 Let f : J R be given by f(x) = ax+b. Then f is differentiable on R with ′ f (c) = a, c2R by using " de nition. Sol. Let c be an arbitrary real number. Consider the expression f(x)f(c) (ax+b)(ac+b) a(xc) = = = a: xc xc xc ′ ′ This suggests that = f (c) = a. Now we will prove that f (c) = a by using " de nition. Let " 0 be given. Now let us try to estimate the error term: jf(x)f(c) (xc)j =ja(xc)a(xc)j = 0: This suggests that we can choose any  0 for any " 0. Let " 0 be given. Let  0 be arbitrary. We can estimate the error term: jf(x)f(c) (xc)j = 0 "jxcj: ′ Since c is arbitrary real number, f is differentiable onR and f (c) = a. x Example 25 Find the derivative of the function f(x), where (i) f(x) = e ; 2 R, (ii) f(x) = log x. a 12 x (x+h) x x h Sol. (i) Given f(x) = e . Then f(x+h)f(x) = e e = e (e 1). Hence h f(x+h)f(x) e 1 x lim = e lim h0 h h0 h 2 1 ( h) x x = e lim h+ + = e : h0 h 2 ( ) 1=h f(x+h)f(x) log (x+h)log x h a a (ii) Given f(x) = log x. Then = = log 1+ . a a h h x Hence ( ) 1=h f(x+h)f(x) h lim = limlog 1+ a h0 h0 h x ( ) 1=h h = log lim 1+ a h0 x 1=x = log e a = (1=x)log e: a Example 26 Discuss the differentiability of the following functions: x; 0 x 1; 2 (i) f(x) = at x = 1 and (ii) f(x) = x on the interval 0;1. 1; x 1: x; 0 x 1; Sol. (i) Given that f(x) = 1; x 1: Now f(x)f(1) x1 D f(1) = lim = lim = 1; x10 x1 x1 x1 f(x)f(1) 11 D f(1) = lim = lim = 0: + x1+0 x1+ x1 x1 ) D f(1)̸= D f(1). + f(x)f(1) ′ Thus f (1) = lim does not exist. Hence f is not differentiable at x = 1. x1 x1 2 (ii) Given that f(x) = x . Let x be any point on (0;1), then 0 2 2 f(x)f(x ) x x 0 ′ 0 f (x ) = lim = lim = lim(x+x ) = 2x : 0 0 0 xx xx xx 0 xx 0 xx 0 0 0 At the end points, we have 2 f(x)f(0) x ′ f (0) = lim = lim = lim x = 0; x0+0 x0+ x0+ x0 x and 2 f(x)f(1) x 1 ′ f (1) = lim = lim = lim x+1 = 2; x10 x1 x1 x1 x1 Thus the function f is derivable in the closed interval 0;1. Theorem 4 (without proof) A function which is differentiable at a point is continuous at the point; but not conversely. 13Theorem 5 (without proof) Let f;g : J R be differentiable at c2 J. Then the following hold: ′ ′ ′ (i) f +g is differentiable at c with (f +g)(c) = f (c)+g (c), ′ ′ (ii) f is differentiable at c with ( f)(c) = f (c), ′ ′ ′ (iii) fg is differentiable at c with (fg)(c) = f (c)g(c)+f(c)g (c), ′ ′ f (c)g(c)f(c)g (c) ′ (iv) f=g is differentiable at c with (f=g)(c) = , provided g(c) = ̸ 0, 2 fg(c)g ′ ′ ′ (v) (gof) is differentiable at c with (gof)(c) = g (f(c))f (c). Example 27 The function f(x) =jxj is continuous on R but not differentiable at x = 0. Sol. Given that f(x) =jxj; 8x2R. Now lim f(x) = lim f(x) = 0 = f(0): x0+ x0 Thus f is continuous at x = 0. We can easily check that f is continuous for all x 0 and x 0. Thus f is continuous for all x2R. But jxj0 x D f(0) = lim = lim = lim 1 = 1: + x0+ x0+ x0+ x0 x and jxj0 x D f(0) = lim = lim = lim (1) =1: x0 x0 x0 x0 x Thus D f(0)̸= D f(0). + Hence the function f is not differentiable at x = 0. xsin(1=x); x = ̸ 0; Example 28 The function f(x) = is continuous but differentiable 0; x = 0; at x = 0. Sol. This function is continuous at x = 0 (see Example-17). But since f(x)f(0) xsin(1=x) = = sin(1=x); x0 x it does not possess left or right limit at x = 0. Hence f is not differentiable at x = 0. 2 x sin(1=x); x̸= 0; ′ Example 29 The function f(x) = is differentiable at x = 0 but f 0; x = 0; ′ ′ is not continuous at x = 0 (i.e., limf (x)̸= f (0)). x0 Sol. We have 2 f(x)f(0) x sin(1=x) ′ Df(0) = f (0) = lim = lim = lim xsin(1=x) = 0: x0 x0 x0 x0 x ′ Thus f differentiable at x = 0 and f (0) = 0. If x̸= 0, then from the elementary calculus, we know that ′ f (x) = 2xsin(1=x)cos(1=x): (4) ′ Clearly, limf (x) does not exist as cos(1=x) oscillates at x = 0 and therefore there is no x0 ′ ′ possibility of limf (x) being equal to f (0). x0 ′ Thus f (x) is not continuous at x = 0. ′ Note further from equation (4) that f is not differentiable at x = 0. 14Assignment 4 Test the differentiability of the following functions at the points indicated. sinx; x =2; (i) f(x) = at the point x = =2 2 1+(x=2) ; x =2; 2 1=(x 1) e ; jxj 1; (ii) f(x) = at the points x =1 and x = 1: 0; jxj 1; 2 1=(x 1) e ; x 1; (iii) f(x) = at the point x = 1: 0; x 1; (iv) f(x) = x(x1) at x = 0 and x = 1. (v) f(x) =jxj+jx1j at x = 0 and x = 1. 2 1=x e ; x 0; (vi) f(x) = at the point x = 0: 0; x 0; 2x3; 0 x 2; (vii) f(x) = at the points x = 2 and x = 4: 2 x 3; 2 x 4; 8 1=x x(e 1) ; x = ̸ 0; 1=x (viii) f(x) = at the point x = 0: e +1 : 0; x = 0; 8 1=x 1=x x(e e ) ; x̸= 0; 1=x 1=x (ix) f(x) = at the point x = 0: e +e : 0; x = 0; 1 xtan (1=x); x = ̸ 0; (x) f(x) = at the point x = 0: 0; x = 0; Theorem 6 Darboux Theorem (without proof) ′ ′ Let f : a;b R be differentiable. Assume that f (a)  f (b). Then there exists c 2 ′ ′ (a;b) such that f (c) = . (Thus though f need not be continuous, it enjoys the intermediate value property.) Intermediate value theorem for derivatives- If a function f is derivable on a closed ′ ′ ′ ′ interval a;b and f (a) ̸= f (b) and  a number lying between f (a) and f (b), then there ′ exists at least one point c2 (a;b) such that f (c) = . 2 x sin(1=x); x̸= 0; Example 30 De ne f(x) = 0; x = 0: Then f is differentiable at all points including 0. 2xsin(1=x)cos(1=x); x = ̸ 0; ′ Here f (x) = 0; x = 0: ′ It is easy to see that f is not continuous (see Example-29). ′ According to Darboux theorem, f enjoys the intermediate value property, even though it is not continuous. 4 Mean Value Theorems On the basis of a certin amount of knowledge about the derivative of a function, mean value theorems enable us to get some information about the function itself. Sometimes, it is easier to tackle the derivative than the function. In this section we shall establish the so-called rst derivative test. 15Theorem 7 Rolle's Theorem (without proof) Let f : a;bR be such that (i) f is continuous on a;b, (ii) f is differentiable on (a;b), ′ and (iii) f(a) = f(b). Then there exists at least one real point c2 (a;b) such that f (c) = 0. The geometric interpretation is that there exists at least one point c2 (a;b) such that slope ofthetangenttothegraphoff atcequalstozero. Thatis, thetangentat(c;f(c))isparallel to the x-axis. There is also an algebraic interpretation of Rolle's Theorem. If f(a) = f(b) = 0, then a and b are the zeros of f(x) or the roots of the equation f(x) = 0. Thus, Rolle's Theorem says that if a and b are two roots of the equation f(x) = 0, then there exists at least one root ′ c2 (a;b) of the equation f (x) = 0. 4 Example 31 Show that the equation 10x 6x+1 = 0 has a root between 0 and 1. First determine the polynomial function f whose derivative is the polynomial, whose roots 5 2 are being sought. So we take f(x) = 2x 3x +x. It is easily seen that f is continuous on 0;1 and differentiable on (0;1). Also f(0) = 0 = f(1). ′ 4 HenceusingtheRolle'stheorem, thereexistsc2 (0;1)suchthatf (c) = 10c 6c+1 = 0. 3 Example 32 Prove that the function f(x) = x +x+k = 0; k in any real constant, has exactly one real root. If it has two distinct roots, say c and c , then f(c ) = 0 and f(c ) = 0. 1 2 1 2 Again, any polynomial function is continuous and differentiable on R. Hence by Rolle's ′ 2 theorem there exists c 2 (c ;c ) such that f (c) = 3c +1 = 0, which contradicts the fact 1 2 that c2R. Hence f has exactly one real root on R. Theorem 8 Mean Value Theorem (MVT) or Langrange's Mean Value Theorem (without proof) Let f : a;b R be such that (i) f is continuous on a;b and (ii) f is differentiable on (a;b). Then there exists at least one real number c2 (a;b) such that ′ f(b)f(a) = (ba)f (c): There is a geometric interpretation of Mean Value Theorem. Under the given conditions, there exists c2 (a;b) such that the slope of tangent to the graph of f at c equals that of the chord joining the two points (a;f(a)) and (b;f(b)). Theorem 9 Cauchy's form of Mean Value Theorem (without proof) Let f;g : a;b R be such that (i) f;g are continuous on a;b, (ii) f;g are differentiable ′ on (a;b), and (iii) g (x)̸= 0, for any x2 (a;b). Then there exists at least one real number c2 (a;b) such that ′ f(b)f(a) f (c) = : ′ g(b)g(a) g (c) Geometrically, Cauchy's form of MVT (Mean Value Theorem) means the following: 2 We look at the map t7 (g(t);f(t)) fromJ toR as a parameterized curve in the plane. For example, t7 (cost;sint);t2 0;2 is parameterization of a circle. f(b)f(a) Then the slope of the chord joining the points (g(a);f(a)) and (g(b);f(b)) is . g(b)g(a) ′ ′ The tangent vector to the parameterized curve at a point (g(t );f(t )) is (g (t );f (t )) and 0 0 0 0 16′ ′ hence the tangent line at t has the slope f (t )=g (t ). Thus Cauchy's mean value theorem 0 0 0 ′ ′ says that there exists a point c2 (a;b) such that slope f (c)=g (c) of the tangent to the curve at c is equal to the slope of the chord joining the end points of the curve. As observed in the remark on the geometric interpretation, if g(x) = x in the Cauchy's mean value theorem, it reduces to Lagrange's mean value theorem. Theorem 10 Applicatons of Mean Value Theorem (MVT) Let f : a;bR be differentiable on (a;b). ′ (i) If f (x) 0 for all x2 (a;b), then f is strictly increasing on (a;b), ′ (ii) If f (x) = 0 for all x2 (a;b), then f is a constant on (a;b), ′ (iii) If f (x) 0 for all x2 (a;b), then f is strictly decreasing on (a;b). The mean value theorem is quite useful in proving certain inequalities. Here are some samples. x Example 33 Show that e 1+x for all x2Rnf0g. x x Supposex 0. Considerthefunctionf(x) = e ontheinterval0;x. Sincee isdifferentiable onR,wecanapplymeanvaluetheoremtof ontheinterval0;x. Hencethereexistsc2 (0;x) such that x 0 ′ c e e = f (c)(x0) = e x: ′ x x c Note that f (x) = e 1 for all x 0. So the displayed equation gives e 1 = e x x. x Similarly if x 0, then consider the interval x;0 and we can prove that e 1+x. yx y yx Example 34 Prove that log ; 0 x y. y x x Let 0 x y and f(x) = logx on x;y. We know that logx is differentiable function on x 0. Hence using the MVT, there exists c2 (x;y) such that y 1 ′ logylogx = f (c)(yx) ) log = (yx): x c 1 1 1 Since 0 x c y, we have . Hence we get y c x yx 1 y yx (yx) = log : y c x x sin sin Example 35 Prove that = cot, for 0  =2. cos cos Let f(x) = sinx and g(x) = cosx for x2 ; . ′ ′ ) f (x) = cosx and g (x) =sinx. Herethefunctionsf andg arebothcontinuousanddifferentiable. Therefore, byCauchy's mean value theorem on ; , sin sin cos = ;  ; cos cos sin or, sin sin = cot;  : cos cos 17x Example 36 Prove that log(1+x) x for all x 0. 1+x 1 x ′ Let f(x) = xlog(1+x). Hence f (x) = 1 = 0. So by Theorem 10, f is 1+x 1+x strictly increasing. Since f(0) = 0;f(x) 0 for x 0. Thus x log(1+x). x Similarly we consider the function g(x) = log(1+x) and show that g(x) 0 for 1+x x 0. tanx x Example 37 Show that for 0 x =2. x sinx 2 tanx x sinxtanxx We have to show that 0, or 0 for 0 x =2. Since x sinx xsinx 2 xsinx 0 for 0 x =2, therefore it will sufficient to show that sinxtanxx 0. 2 ′ 2 Let f(x) = sinxtanxx , then for 0 x =2, f (x) = sinx+sinxsec x2x. We ′ cannot decide about the sign of f (x) mainly because of the presence of the (2x) term. Now ′ the function f (x) is continuous and derivable on (0;=2). ′′ 2 2 ) f (x) = cosx+cosxsec x+2sinxsec xtanx2 p p 2 2 = ( secx cosx) +2tan xsecx 0; for 0 x =2: ′′ ′ ′ Since the derivative f (x) of f (x) is positive, the function f (x) is an increasing function. ′ ′ Further since f (0) = 0, therefore the function f (x) 0 for 0 x =2. ′ Again, since f (x) 0, f(x) is an increasing function and because f(0) = 0, the function f(x) 0, for 0 x =2. tanx x Thus it follows that for 0 x =2. x sinx Assignment 5 Solve the following problems by using Rolle's theorem/ MVT; x 1. Prove that between any two real roots of e sinx = 1, there is at least one real root of x e cosx+1 = 0. 3 2 2. Show that the equation cosx = x +x +4x has exactly one root in 0;=2. 3 2 3. Prove that the equation x 3x +b = 0; b2R has at most one r 4. Let f : 2;5R be ′ 2 continuous and be differentiable on (2;5). Assume that f (x) = (f(x)) + for all x2 (2;5). True or false (with proper reason): f(5)f(2) = 3. 5. Apply Lagrange's mean value theorem for the function f(x) = log(1+x) to show that 1 1 0 1; for all x 0: log(1+x) x 6. Establish the following inequalities: 2 3 2 3 x x x x (i) x + log(1+x) x + ; x 0, 2 3(1+x) 2 3 2 2 x x (ii) xlog(1+x) ; x 0, 2(1+x) 2 2 2 x x (iii) xlog(1+x) ; 1 x 0, 2 2(1+x) x x (iv) (1+x) e 1+xe ; for all x, 2 x x (v) (1x) e 1x+ ; for all x 0. 2 18Theorem 11 Taylor's Theorem (without proof) (n) (n+1) Assume that f : a;b R be such that f is continuous on a;b and f (x) exists on (a;b). Fix x 2 a;b. Then for each x2 a;b with x = ̸ x , there exists c between x and x 0 0 0 such that n k ∑ (xx ) 0 (k) f(x) = f(x )+ f (x )+R ; (5) 0 0 n k k=1 n+1 (xx ) 0 (n+1) where R = f (c). n (n+1) The right-hand side of (5) is called the n-th order (or n-th degree) Taylor expansion of n k ∑ (xx ) 0 (k) the function f at x . The expression f(x) = f(x ) + f (x ), is called the 0 0 0 k k=1 n-th degree Taylor polynomial of f at x . The term R is called the remainder term in the 0 n Taylor's expansion after (n+1) terms. The remainder term is the \error term" if we wish (n+1) to approximate f near x by the n-th order Taylor polynomial. If we assume that f is 0 n bounded, say, by M on (a;b), then R goes to zero much faster than (xx ) 0. n 0 Putting x = 0 in expression in (5) is called Maclaurin's Theorem with form of remainder. 0 Example 38 Show that tha function f(x) = sinx; x2 0;=4 is approximated by a polyno- 3 mial sinx = x(x =6), with an error less than 1=400. The function f satis es the condition of Taylor's theorem. Hence it can be expressed at x = 0 by 0 2 3 4 x x x ′ ′′ ′′′ iv f(x) = f(0)+xf (0)+ f (0)+ f (0)+ f (0)+R ; 5 2 3 4 5 x (v) where for some c2 (0;x), and R = f (c). Thus 5 5 3 5 x x f(x) = x + cosc: (6) 6 120 Now 5 5 x x 1 1 5 jR j = jcoscj  (=4) : (7) 5 120 120 120 400 Hence it follow from (6) and (7) that 3 1 x jf(x)p(x)j ; where p(x) = x : 400 6 Assignment 6 1. If 0 x 2, then prove that 2 3 4 (x1) (x1) (x1) logx = (x1) + + : 2 3 4 2. Assuming the validity of expansion, show that 3 2 4 2 5 2x 2 x 2 x x (i) e cosx = 1+x + : 3 4 5 1 1 2 4 (ii) logsecx = x + x + : 2 12 2  x=4 (x=4) 1 1 (iii) tan x = tan + + : 2 2 2 4 1+ =16 4(1+ =16) ( ) 2 3  1   (iv) sin( +) = p 1+ + : 4 2 3 2 19References 1 A. Kumar and S. Kumaresan, A basic course in Real Analysis, CRC Press, 2014. 2 G. Das and S. Pattanayak, Fundamentals of Mathematical Analysis, Tata McGraw Hill Publication, 2010. 3 S.C. Malik and S. Arora, Mathematical Analysis, New Age International Publication, 2008. 20

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