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- 1. Limits, Continuity & Differentiation
- 2. LimitsLimits
- 3. Concepts 3 Points to remember
- 4. Meaning of x approaches to a : An independent variable x approaches (tending) to a is denoted by When we write , we shall assume the following: x a→ x a→ 4 When we write , we shall assume the following: (i) x is a real variable. (ii) A is a finite real number whose value is fixed. (iii) x passes successively through a number of values so that if we consider any particular value taken by x , we may discriminate the values that precede and the values that follow and none of the values can be looked upon as the last. x a→
- 5. Definition-1 : Or lim x = a means: Given any (no matter how small) the successive values of x will x a→ 0ε> 5 Given any (no matter how small) the successive values of x will ultimately satisfy the inequality i.e ultimately x lies in the open interval i.e ultimately x satisfies and 0ε> 0 x a< − <ε ( )a ,a ;x a−ε + ε ≠ a x a−ε < < a x a< < + ε
- 6. When we say that x tends to a from the left and we write Or but when we say that x tends to a from the right and we write or a x a :−ε < < x a or x a− → − → 0 x a− → a x a ;< < + ε x a→ +tends to a from the right and we write orx a→ + x a 0 or x a+ → + → 6 Remember 0 0 x a x a x a− + → ⇒ → ∧ →
- 7. Meaning of x approaches to : An independent variable x approaches (tending) to is denoted by When we write , we shall assume the following: (i) x is a real variable. which assumes successive values so that x → +∞ x → +∞ +∞ ∞ 7 (i) x is a real variable. which assumes successive values so that one can discriminate values that precede and the values that follow and that no value is the last value attained by x (ii) is not a number at all; it is a symbol whose meaning will be made clear in the following lines now means : the successive values of x ultimately become and remain more than any arbitrary positive number G (no matter how large) given in advance +∞ x → ∞
- 8. Meaning of x approaches to : An independent variable x approaches (tending) to is denoted by When we write , we shall assume the following: (i) x is a real variable. x→−∞ x →−∞ −∞ −∞ 8 (i) x is a real variable. (ii) is same as now means : the successive values of x ultimately become and remain less than −G (where G is any arbitrary positive number) given in advance. x →−∞ ( x)− → +∞ x →−∞
- 9. Limit of a function at a point: A function f(x) is said to tend to a limit l as x tends to a if to each given , there exists a positive number (depending on ) such that0ε > δ εgiven , there exists a positive number (depending on ) such that | f (x) l | whenever 0 | x a |ε δ− < < − < i.e,. for all those values of x (except at x =a) which belongs to . This is denoted by ( ) ( , )f x l lε ε∈ − + ( , )a aδ δ− + lim ( ) x a f x l → = 9
- 10. Left-hand and right-hand limits f(x) is said to tend to l as x tends to a through values less than a, if to each such that0, 0,ε δ> ∃ > | ( ) | whenf x l a x aε δ− < − < < So that whenever The limit in this case is called the left-hand limit (L.H.L) and is denoted by f(a-0). Thus 10 | ( ) | whenf x l a x aε δ− < − < < ( ) ( , )f x l lε ε∈ − + ( , )x a aδ∈ − 0 ( 0) lim ( ) x a f a f x → − − =
- 11. Similarly, if f(x) tends to l as x tends a through values which are greater than a i.e. if given such that0, 0ε δ> ∃ > | ( ) | when ,f x l a x aε δ− < < < + then f(x) is said to tend to l from the right and the limit so obtained is called the right hand limit (R.H.L) and is denoted by f(a+0). we write 11 | ( ) | when ,f x l a x aε δ− < < < + 0 ( 0) lim ( ) x a f a f x → + + =
- 12. Existence of a limit at a point. f(x) is said to tend to a limit as x tends to ‘a’ if both the left and right hand limits exist and are equal, and their common value is called the limit of the functionlimit of the function i) To find f(a-0) or we first put x = a-h, h>0 in f(x) and then take the limit as Thus ii) To find f(a+0) or we first put x = a+h, h>0 in f(x) and then take the limit as Thus 0 lim ( ) x a f x → − 0h → + 0 0 lim ( ) lim ( ) x a h f x f a h → − → + = − 0 lim ( ) x a f x → + 0h → + 0 0 lim ( ) lim ( ) x a h f x f a h → + → + = + 12 How to Find L.H.L & R.H.L ?
- 13. Limits at Infinity And Infinite Limits i) A function f(x) is said to tend to l as if given however small, number k (depending on ) s.t lim ( ) x f x l →∞ = x → ∞ 0,ε > a ve∃ + ε | ( ) | . ., 1 ( ) .f x l x k i e f x l x kε ε ε− < ∀ ≥ − < < + ∀ ≥ ii) A function f(x) is said to tend to l as if given however small, number k(depending on ) s.t | ( ) | . ., 1 ( ) .f x l x k i e f x l x kε ε ε− < ∀ ≥ − < < + ∀ ≥ lim ( ) x f x l →−∞ = x → −∞ 0,ε > a ve∃ + ε | ( ) | . ., ( )f x l x k i e l f x l x kε ε ε− < ∀ ≤ − − < < + ∀ ≤ − 13
- 14. iii) A function f(x) is said to tend to as x tends to a, if given k>0, however large, number s.t. ∞ a+ve∃ δ lim ( ) x a f x → = ∞ iv) A function f(x) is said to tend to as x tend to a, if given k>0, however large, number s.t ( ) for 0 | |f x k x a δ> < − < − ∞ a+ve∃ δ lim ( ) x a f x → = −∞ ( ) for 0 | |f x k x a δ< − < − < 14
- 15. v) A function f(x) is said to tend to as , if given k>0, however large, number s.t. ∞ a∃ ( )f x k x k′> ∀ ≥ lim ( ) x f x →∞ = ∞ x → ∞ k 0′> vi) A function f(x) is said to tend to as , if given k>0, however large, number s.t ( )f x k x k′> ∀ ≥ − ∞ a∃ lim ( ) x f x →∞ = −∞ ( )f x k x k′< − ∀ ≥ 15 x → ∞ 0k′ >
- 16. vii) A function f(x) is said to tend to as , if given k>0, however large, number s.t. ∞ a∃ ( )f x k x k′> ∀ ≤ − lim ( ) x f x →−∞ = ∞ x → −∞ k 0′> viii) A function f(x) is said to tend to as , if given k>0, however large, number s.t ( )f x k x k′> ∀ ≤ − − ∞ a∃ lim ( ) x f x →−∞ = −∞ ( )f x k x k′< − ∀ ≤ − 16 x → −∞ 0k′ >
- 17. So From the above all definitions we conclude that there are four types in limits at finite point 17 at finite point at infinite point lim value finite value infinite
- 18. The existence of limit l of f(x) as is illustrated in the figure l and being known, one can draw the horizontal lines if only a of the point x =a can be found s.t. all points of the graph of y =f(x) corresponding to points x of the nbd (excepting perhaps Limit of the Function Geometric Point of View x a→ ε y l,y l and y l ,y l as x a= = −ε = +ε → → nbdδ− y = f(x)to points x of the nbd (excepting perhaps for the point x=a) lie within the strip bounded by the line The point P(a,l) may or may not belong to the graph of y = f(x). We are simply not concerned with it. 18 andy l x a= ±ε = ±δ y = f(x) y = f(x) y = l +ε y = l -ε y = l a a + δa -δ p(a,l)
- 19. Algebra & Properties of LimitsAlgebra & Properties of Limits
- 20. Indeterminate forms If a function f(x) takes any of the following forms at ;x a= 0 00 , , ,0 ,0 , ,1 0 ∞∞ ∞ − ∞ × ∞ ∞ ∞ then f(x) is said to be indeterminate at .x a= Notes:Notes: 1. All indeterminate forms (only limiting value) can be calculated. 2. Infinity (∞ ) is a symbol and not a number and we can not draw ∞ on the paper. 3. Zero (0) is signless quantity i.e., neither positive nor negative and 0 is even. 20
- 21. 4. (i) ∞ + ∞ = ∞ (ii) ∞ × ∞ = ∞ (iii) ∞ ∞ = ∞ (iv) 0 0∞ = (v) 0 ∞ = ∞ (vi) 0 0= ∞ (vii) 0 0= ∞ where ais finite a (viii) 0 a is undefined, if 0a ≠ (ix) 0 0ab a= ⇔ = or 0b = or 0a = , 0b = ,a and b are finite. (x) , (so not exists) if a>1 1, if a=1 0, if -1<a<1 not exists if a -1 a∞ ∞ = ≤ (xi) a ,a ,a , ,0 ,a+ ∞ = ∞ ×∞ = ∞ − ∞ = −∞ ∞ − = ∞ − ∞ = −∞ 0 ,+ ∞ = ∞ where a is a non-zero positive quantity. 21
- 22. Property 1 If f(x) is an even function, then ( ) ( )0 0 lim lim x x f x f x → − → + = Property 2 If f(x) is an odd function and ( )lim f x exists, then. ( )lim 0f x =If f(x) is an odd function and ( )0 lim x f x → exists, then. ( )0 lim 0 x f x → = Notes: If a function involves square root function, modulus function, greatest integer function, least integer function, fractional part function, signum function etc. then limit of this function is likely to be two sided limit. What is two sided limit? 22
- 23. 1. If ( ) ( ) ( )lim lim lim x a x a x a f x l f x f x l → → − → + = ⇔ = = where lis finite and unique. 2. ( ) 0 1 lim lim x x f x f x→±∞ → = 3. (i) 0 1 lim x x→ + = +∞ and 0 1 lim x x→ − = − ∞ 0 1 lim x x→ ∴ does not exist. 0x x→ + 0x x→ − 0x x→ X Y O0x→ − 0x→ + 23
- 24. (ii) 2 20 0 1 1 lim lim x xx x→ + → − ∴ = + ∞ = 20 1 lim x x→ ∴ does not exists. Since ∞ is a symbol, not a number which is not finite. Y X O 24
- 25. Difference between ( )lim x a f x → and ( )f a Let f(x) be a function and a be a point then following cases arise: (i) ( )lim x a f x → exists but f(a) does not exist: Consider the function ( ) 2 2 x a f x x a − = −x a− ∴ The value of function f(x) at x a= is of the form 0 0 which is indeterminate. i.e., ( ) 0 0 f a = ∴ f(a) does not exist. 25
- 26. But ( ) ( ) ( ) ( ) ( ) 2 2 0 0 0 lim lim lim lim 2 2 x a h h h a h a f x f a h a h a a h a→ − → → → − − = − = = − = − − and ( ) ( ) ( ) ( ) 2 2 0 0 0 lim lim lim lim 2 2 x a h h h a h a f x f a h a h a a h a→ + → → → + − = + = = + = + −0 0 0x a h h ha h a→ + → → →+ − ( ) ( )lim lim x a x a f x f x− + → → ∴ = ∴ ( )lim x a f x → exists. Hence ( )lim x a f x → exist but f(a) does not exist. 26
- 27. (ii) ( )lim x a f x → does not exist but f(a) exists: (iii) ( )lim f x and ( )f a both exists and are equal:(iii) ( )lim x a f x → and ( )f a both exists and are equal: (iv) ( )lim x a f x → and ( )f a both exist but are unequal: 27
- 28. Standard Theorems on Limits Let f be a function defined in a certain interval and containing ‘a’ as interior point . If ( )lim x a f x l → = and ( )lim , x a g x m → = where l and m exist and finite, then (1) ( ) ( ){ } ( ) ( )lim lim limf x g x f x g x l m+ = + = +(1) ( ) ( ){ } ( ) ( )lim lim lim x a x a x a f x g x f x g x l m → → → + = + = + (2) ( ) ( ){ } ( ) ( )lim lim lim x a x a x a f x g x f x g x l m → → → − = − = − (3) ( ) ( ){ } ( )( ) ( )( )lim . lim lim . x a x a x a f x g x f x g x l m → → → = = (4) ( ) ( ) ( ){ } ( ){ } lim lim , lim x a x a x a f xf x l g x mg x → → → = = provided 0m ≠ 28
- 29. (5) ( ){ } ( ) ( ){ } ( )lim lim lim g x m x a x a x a g x f x f x l → → → = = (6) ( ){ } ( )lim lim x a x a kf x k f x kl → → = = (7) If ( ) ( )f x g x≤ for all x then ( ) ( )lim limf x g x l m≤ ⇒ ≤(7) If ( ) ( )f x g x≤ for all x then ( ) ( )lim lim x a x a f x g x l m → → ≤ ⇒ ≤ (8) ( ) ( )lim lim x a x a f x f x l → → = = (9) ( ) ( )lim lim f x l x a x a f x e e e → → = = 29
- 30. (10) ( ) ( ){ }lim lim ( ), x a x a ln f x ln f x ln l → → = = only if 0l > (11) ( )( ) ( ){ } ( )lim lim x a x a f g x f g x f m → → ⇔ = provided f is continuous at x m= (12) Sandwitch / Squeeze theorem: y If ( ) ( ) ( )f x g x h x≤ ≤ ( )δ, δx a a∀ ∈ − + and ( ) ( )lim lim , x a x a f x l h x → → = = then lim x a g( x ) l → = Corollary: If ( ) ( )<f x g x in a certain nbd of a and lim ( ) 0 → = x a g x , then lim ( ) 0 → = x a f x . h(x) f(x) y = g(x) 0 x y 30
- 31. (13) NEIGHBOURHOOD PROPERTY: If , lim ( ) → = x a f x l then there exists some neighbourhood of a , at every point of which f(x) will have the same sign as that of l. Corollary: Let lim ( ) =f x l , Suppose f(x) > 0 for every x in a certain nbd of a then l is not negative .Let lim ( ) → = x a f x l , Suppose f(x) > 0 for every x in a certain nbd of a then l is not negative . Similarly when f(x) < 0 for every x in a certain nbd of a then l is not positive. Also remember that even when f(x) > 0 for all x in a certain nbd of a, still lim ( ) →x a f x may be 0. Eg: f(x) = x 2 If f(x) < g(x) for a – h < x < a + h and x a x a limf (x) l andlimg(x) m → → = = then ≤l m .. 31
- 32. (14) CAUCHY’S CRITERION FOR THE EXISTANCE OF LIMIT OF A FUNCTION: The necessary and sufficient condition that a function f(x) may tend to a definite finite limit l as →x a , is that , to any pre-assigned positive quantity ε , however small , there corresponds a positive number δ , such that 2 1( ) ( )− <εf x f x for every pair x1, x2 of values of x satisfying 0 < − < δx a and 0 < − < δx a .values of x satisfying 10 < − < δx a and 20 < − < δx a . And for limits when → ∞x , The necessary and sufficient condition that a function f(x) may tend to a definite finite limit l as → ∞x , is that , to any pre-assigned positive quantity ε , however small , there corresponds a positive number δ , such that 2 1( ) ( )− <εf x f x for every pair x1, x2 exceed δ 32
- 33. (15) LIMIT OF A FUNCTION DEFINED IN TERMS OF SEQUENCE: Let f be defined in some nbd of a Definition-1:- lim ( ) 0, 0 ( ) → = ⇒ ∈ > ∃ δ > − < ∈ x a f x l G iven a n y su ch th a t f x l( ) 0 − < ∈ ∈ < − < δ f x l fo r a ll x x a Definition-2:- { } { } lim ( ) lim ( ) → = ⇒ → → ∞ n x a n f x l it of thesequence f x l as n for all sequence x (in the nbd of a where f is defined) which converge to a. Remember that the above two definitions are equivalent. 33
- 34. Conceptual Problems EXAMPLE-1: Let 1, ( ) 0, c if x Q f x if x Q ∈ = ∈ Prove that lim ( ) x a f x → does not exist where ‘a’ is any real number. Solution:Solution: Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let { }ny be a sequence of irrational numbers such that ny a→ and ny a≠ , n=1,2,3…..;.Clearly ( ) 1, ( ) 0n nf x f y= = for n=1,2,3…… Therefore ( ) 1nf x → and ( ) 0nf y → and lim ( ) x a f x → does not exist. Since ‘a’ is arbitrary, limit does not exist for every real number. 34
- 35. EXAMPLE-2: Let , ( ) 0, c x if x Q f x if x Q ∈ = ∈ Prove that lim ( ) x a f x → exists only if a=0 . Solution: suppose a ≠ 0. { } { }Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let { }ny be a sequence of irrational numbers such that ny a→ and ny a≠ , n=1,2,3…..;.Clearly ( ) , ( ) 0n n nf x x f y= = for n=1,2,3…… Therefore ( )n nf x x a= → and ( ) 0nf y → and lim ( ) x a f x → does not exist. Since ‘a’ is arbitrary, limit does not exist for every non-zero real number. Now for any sequence { }nx , ( )n nf x x= or 0 and so if 0nx → then ( ) 0nf x → .Therefore , by above definition 0 lim ( ) 0 x f x → = 35
- 36. Important Expansions (i) ( ) ( ) ( )( )2 31 1 2 1 1 ... 1.2 1.2.3 n n n n n n x nx x x − − − + = + + + + (ii) 2 3 1 ... 1! 2! 3! x x x x e = + + + + 1! 2! 3! (iii) ( ) ( ) 2 3 3 1 ... 1! 2! 3! x xlna x x a lna lna 2 = + + + + (iv) ( ) 2 3 4 1 ...; 1 1 2 3 4 x x x ln x x x+ = − + − + − ≤ ≤ 36
- 37. (v) 1 2 2 3 1 . ... n n n n n nx a x ax a x a x a − − − −− = + + + + − (vi) 3 5 sin ... 3! 5! x x x x= − + − (vii) 2 4 cos 1 ... 2! 4! x x x = − + − (viii) 3 52 tan ... 3 15 x x x x= + + + (ix) 2 3 2 2 5 2 2 2 1 71 . 1 .3 . 1 .3 .5 sin . ... 3! 5! 7! x x x x x− = + + + + 37
- 38. (x) 3 5 1 tan 3 5 x x x x− = − + − (xi) 2 4 6 1 5 61 sec 1 ... 2! 4! 6! x x x x− = + + + + 2! 4! 6! (xii) ( ) 2 2 2 21 2 4 62 2.2 2.2 .4 sin ... 2! 4! 6! x x x x− = + + + (xiii) ( ) 1/ 211 1 1 ... 2 24 x x x e x + = − + + 38
- 39. Standard Limits & Short cuts in Limits If ( ) 0f x → when ,x a→ then (1) ( ) ( ) 1 lim 1 f x x a e f x→ − = (2) ( ) ( ) ( ) 1 lim , 0 f x c ln c c f x→ − = >(2) ( ) ( )lim , 0 x a ln c c f x→ = > (3) ( )( ) ( ) 1 lim 1 x a ln f x f x→ + = (4) ( ) ( ) sin lim 1 x a f x f x→ = (5) ( ) ( ) tan lim 1 x a f x f x→ = 39
- 40. (6) ( ){ } ( ){ } ( ) ( ) ( ){ } 1 lim , n n n x a f x f a n f a n Q f x f a − → − = ∈ − (7) ( ){ } ( )1/ lim 1 f x x a f x e → + = (8) ( ){ } ( )/ lim 1 c f x bc bf x e+ =(8) ( ){ } ( )/ lim 1 c f x bc x a bf x e → + = (9) ( ) 1/ 0 1 lim 1 lim 1 n h n h e h n→∞ → + = = + (10) lim 1 n a n a e n→∞ + = 40
- 41. (11) ( ) 1/ 0 lim 1 h a h ah e → + = (12) 1 lim n n n x a x a na x a − → − = − (13) lim m m m nx a m a −− =(13) lim m n n nx a x a m a x a n − → − = − (14) 0 1 lim 1 x x e x→ − = (15) ( )0 1 lim , 0 x x a ln a a x→ − = > (16) ( ) 0 1 lim 1 x ln x x→ + = 41
- 42. (17) ( ) ( )0 log 1 lim log 0, 1a a x x e a a x→ + = > ≠ (18) 1 1 0 0 0 0 sin tan sin tan lim 1 lim lim lim x x x x x x x x x x x x − − → → → → = = = = 0 π (where x is measured in radians) and ( ) 0 0 sin lim 180 180 c x x x π π → = ° = ° (19) ( )lim 0 0mx ln x m x→∞ = > (20) ( ) 0 1 1 lim m x x m x→ + − = 42
- 43. (21) ( ){ } ( ) ( ){ } ( )lim 1 lim x a x ag x f x g x f x e → → − = This is Very Very Important Formula in Limits provided ( ) ( ), 1g x f x→ ∞ → for x a→ 43
- 44. (22) If ( )lim 0 x a f x α → = > and ( )lim x a g x β → = (a finite quantity) then ( ){ } ( ) β lim α g x x a f x → = n 2 1-cos ax na (23). n 2 2x 0 1-cos ax na lim x 2→ = (24). ( )2 2n n 2x 0 n b acos ax cos bx lim x 2→ −− = (25). n n n 2 n 2x 0 tan x sin x na lim x 2 + +→ − = 44
- 45. (26). b/x ab x 0 lim (1 ax) e→ + = (27). bx ab x 0 a lim 1 e x→ + = (28). d/x bcd lim(cosax bsincx) e+ =(28). d/x bcd x 0 lim(cosax bsincx) e→ + = (29). d/x bcd x 0 lim (secax b tancx) e→ + = (30). b/xx x x x 1 2 3 n x 0 a a a a lim n→ + + +−−−−+ = (a1a2------an)b/n (31). ( )2 x 0 a lim x ax b x 2→ + + − = 45
- 46. (32). ∑ np is the (p+1)th degree polynomial function in the variable n and in the expansion. ∑ np , the coefficient of np+1 is 1 1 +p (33). x f (x) lim g(x)→∞ , when f(x) & g(x) are the polynomial functions in x of degrees m & n respectively. Thenrespectively. Then I. if m<n, then x f (x) lim 0 g(x)→∞ = II. if m=n, then n nx f (x) coeff.of x in f(x) lim g(x) coeff.of x in g(x)→∞ = III. If m>n, then x f(x) lim g(x)→∞ =∞, if coefficient of xm in f(x) is +ve = -∞, if coefficient of xm in f(x) is –ve 46
- 47. (34). 1 1 1 0 − = −+ → n m nn m a nx axa x Lt (35). 1 1 1 0 − = −− → n m n mn a nx xaa x Lt 1 2−−+ n mn m xaxaLt (36). 1 1 2 0 − = −−+ → n m n mn m a nx xaxa x Lt (37). ( )21 2 1 aab cbx e ax ax x Lt − + = + + ∞→ (38). ( )21 22 2 11 2 0 aac dcx e bxax bxax x Lt − + = ++ ++ → 47
- 48. (39). ( )21 2 2 2 1 2 0 bbc edxcx e baxx baxx x Lt − ++ = ++ ++ → (40). ( )functionintegergreatestthedenotes[.] ][ a x bax x Lt = + ∞→ 1 1 1Lt (41). 1 1 1 terms 2 Lt n n a( a d ) ( a d )( a d ) ad + +−−−−− = → ∞ + + + (42). [ ] [ ] [ ] [ ] 1 321 1 + = +−−−−−+++ ∞→ + p x n xnxxx n Lt p pppp where [.]denoted GIF Remember 1) ( )0 lim 1 0 n h n h → →∞ − = 2) ( )0 lim 1 n h n h → →∞ + = ∞ 48
- 49. Methods For Finding Limit of the Function Method-1: Direct Substitution Method Directly substitute the point in the given expression if on substitution the expression does not take indeterminate form and if on substitution we get a finite number, then the finite number is the limit of the given expression. Method-2: Factorization Method ( )f x Consider ( ) ( ) lim x a f x g x→ . If we put the value x a= the rational function ( ) ( ) f x g x takes the meaningless forms 0 , 0 ∞ ∞ etc. then ( )x a− is a factor of both ( )f x and ( )g x which must be cancelled. After cancelling out the common factor ( )x a− we again put x=a in the given expression then we get a meaningful number or not. We have to repeat the above process till we get a meaningful number 49
- 50. Note: 1. ( )( )1 2 3 2 4 3 2 1 ...n n n n n n n n x a x a x x a x a x a a x a− − − − − − − = − + + + + + + Where n is even or odd positive integer. 2. ( ) ( )( )11 2 3 2 4 3 1 ... 1 nn n n n n n n x a x a x x a x a x a a −− − − − − + = + − + − + + −2. ( ) ( )( )11 2 3 2 4 3 1 ... 1 nn n n n n n n x a x a x x a x a x a a −− − − − − + = + − + − + + − Where n is odd positive integer. Also this formulae is not applicable when n is even. 50
- 51. 3. Synthetic Division Method: To find the quotient when a polynomial is divisible by a binomial. Let ( ) 1 2 0 1 2 1.....n n n n nf x a x a x a x a x a− − −= + + + + + be a polynomial of degree n and let it is divisible by the binomial x-h. If 1 2 3 4 ...n n n n Q b x b x b x b x b b− − − − −= + + + + + + be the quotient and hereIf 1 2 3 12 ... xo nn Q b x b x b x b x b b −− = + + + + + + be the quotient and here remainder = 0, then the coefficient of Q can be found from the following table: 0a 0 0b 1a 0hb 1b 2a 1hb 2b 3b 2hb 3a 1na − 2nhb− 1nb − ...... ...... ...... na 1nhb − 0 h Where 0 0 1 1 0 2 2 1; ; ;b a b a hb b a hb= = + = + 3 3 2 1 1 2;.....; n n nb a hb b a hb− − −= + = + and 1 0n na hb −+ = 51
- 52. Method-3: Rationalization or Double rationalization Method This method is used when either numerator or denominator or both have fractional powers 1 1 1 like , , 2 3 4 etc .After rationalization of the terms and cancelling common factors in numerator and in denominator which gives the result.numerator and in denominator which gives the result. Method-4: Standard limit Method ( ) 1 lim where ‘n’ is a rational number n n n x a x a n a x a − → − = − 52
- 53. Method-5: Limit at ∞ Step I: Write down the given expression in the form ( ) ( ) f x g x (where ( ) 1g x ≠ ), if ( ) 1g x = then multiplying above and below by conjugate of f(x). Step II: Then divide each term of the numerator and denominator by k x where k is highest power of x in numerator and denominator. Step III: Then use the result lim 0kx c x→∞ = where c is a constant and 0k > . Note: If then then | | and f then | |x x x i x x x→ ∞ = → −∞ = − 53
- 54. Method-6: Exponential and Logarithmic Limits To evaluate the exponential and logarithmic limits we use the following results: (i) 0 1 lim 1 x x e x→ − = 0 form 0 x 0 (ii) 0 1 lim x x a ln a x→ − = 0 form 0 (iii) ( ) 0 1 lim 1 x ln x x→ + = 0 form 0 54
- 55. Method-7:Exponential limits of the form 1∞ To evaluate the exponential limits of the form 1∞ we use the following results: (i) ( ) 1/ 0 1 lim 1 lim 1 n h n h e h n→∞ → + = = + ( ) 1/ n ha (ii) ( ) 1/ 0 lim 1 lim 1 ha n h a e ah n→∞ → + = = + (iii) ( ){ } ( ) ( ){ } ( )lim 1 lim x a f x g xg x x a f x e → − → = provided ( ) ( ), 1g x f x→ ∞ → for x a→ Proof: Let ( ){ } ( ) lim g x x a P f x → = ( )( ){ } ( ) lim 1 1 g x x a f x → = + − ( )( )( ) ( ) ( ) ( )1 1/ 1 lim 1 1 f x g x f x x a f x − − → = + − ( ) ( )lim 1 x a f x g x e → − = 55
- 56. Method-8: ’ L’ Hospital’s Rule If ( )f x and ( )g x be two functions, such that ( ) ( ) 0f a g a= = and f and g are both differentiable at everywhere in some neighbourhood of point a except possibly ‘a’.differentiable at everywhere in some neighbourhood of point a except possibly ‘a’. Then ( ) ( ) ( ) ( ) 1 1 lim lim x a x a f x f x g x g x→ → = Provided ( )1 f a and ( )1 g a are not both zero 56
- 57. Note : 1. L’ Hospital’s rule is applicable only for forms 0 0 or ∞ ∞ . 2. For other indeterminate forms we have to convert to 0 0 or ∞ ∞ and then apply L’ Hospital’s rule. Sometimes we have to repeat the process if the form is 0 0 or ∞ ∞ again. 0 ∞ 3. Method to convert other indeterminate form to either 0/0 or ∞ /∞ form. (i) 1 1 0 0 0 0 0 0.0 0 − ∞ − ∞ = − = = form, in this case take L.C.M. (ii) 1 1 1 0 0 or or 0 0 0 ∞ ×∞ = × ×∞ = ∞ ∞ form (iii) log 1 log 1 0 / 0/0 1 ore e e e e e e ∞ ∞∞ ∞× ∞ ∞ = = = = form (iv) 0 log 0 0log 00 0 0/0 / 0 ore e e e e e e×∞ ∞ ∞ = = = = (v) 0 log 0log0 0 0/0 / ore e e e e e e∞ ∞ ×∞ ∞ ∞ ∞ = = = = 57
- 58. 4. , 1 log 0 , 0 1 a a a −∞ > = ∞ < < and , 1 log , 0 1 a a a ∞ > ∞ = −∞ < < 5. Newton-Leibnitz’s formula: If ( ) ( ) ( ) , x x y f t dt ψ φ = ∫ then { } { }1 1 ( ) ( ) ( ) ( ) dy f x x f x x dx ψ ψ φ φ= × − × This is Very Very Important Formula in Limits 58
- 59. ContinuityContinuity
- 60. Concepts 60 Points to remember
- 61. CONTINUITY AT A POINT AND AN INTERVAL I Continuity at a point: Let f be defined on an interval I R⊆ . Suppose ‘c’ is an interior point of I. The continuity of f at x=c may be defined in the following way: Definition-1:-Definition-1:- f is said to be continuous at x = c if for any arbitrary positive number ε , no matter how small, ∃a positive numberδ such that for all points x x c I∈ − < δ∩ we have ( ) ( )f x f c− < ε .It means if f is continuous at ‘c’ thenlim ( ) x c f x → exists and is equal to f(c). That is ‘f’ is continuous at x = c if the functional values f(x) are close to f(c) when x is close to c .Thus for continuity of f at x = c, f must be defined at x = c and also in some nbd around c; otherwise we cannot find lim ( ) x c f x → which must exist and equals to f(c). 61
- 62. Hence for continuity of ‘f’ at a point x = c , we must have 1) ‘f’ is defined at x = c i.e., f(c) exists. 2) lim ( ) x c f x → exists i.e., lim ( ) lim ( ) x c x c f x f x → + → − = 3) f(c) and lim ( ) x c f x → must be equal. If any one of the above conditions fail, then f is NOT continuous or discontinuous at x=c. Definition-2:- f is said to be continuous at x=c if every convergent sequence{ }nx in I having limit c, the sequence { }( )nf x is convergent and lim ( ) ( )n n f x f c →∞ = . And if f is discontinuous at c iff there exist { }nx in I such that { }nx converges to c but { }( )nf x does not converges to f(c). Remember that definition-1 is equivalent to definition-2. 62
- 63. Hence for continuity of ‘f’ at a point x = c , we must have 1) ‘f’ is defined at x = c i.e., f(c) exists. 2) lim ( ) x c f x → exists i.e., lim ( ) lim ( ) x c x c f x f x → + → − = x c→ x c x c→ + → − 3) f(c) and lim ( ) x c f x → must be equal. If any one of the above conditions fail, then f is NOT continuous or discontinuous at x=c. 63
- 64. Continuity in an Open Interval A function f is said to be continuous in an open interval (a,b) if fi is continuous at every point of (a,b). Thus f is continuous in the open interval (a, b) iff for every limx c c ( a,b ), f ( x ) f (c )→ ∈ = 64
- 65. Continuity in a Closed Interval A function f is said to be continuous in a closed interval [a, b] if it is i) right continuous at a i.e., lim ( ) ( )x a f x f a→ + = ii) Continuous in (a, b) iii) Left continuous at b lim ( ) ( ) x b f x f b → − = 65
- 66. Continuity on a Set A function f is said to be continuous on an arbitrary set ( )S R⊂ if for each 0ε < and for every a S∈ , ∃ a real number 0δ > such that | ( ) ( ) |f x f a− <ε whenever x S∈ and | |x a− <δ Equivalently, a function f is said to be continuous on a set S if it continuous at every point S, i.e. If for every ,lim ( ) ( ) x a a S f x f a → ∈ = 66
- 67. Remember It should be noted that continuity of a function is the property of interval and is meaningful at x = a only if the function has a graph in the immediate neighbourhood of x = a, not necessarily at x = a. Hence it should not be mislead that continuity of a function is talked only in its domain. Example:- Continuity of 1 f (x) x 1 = − at x = 1 is meaningful but continuity of f (x) = ln x at x = – 2 in meaningless. Similarly if f (x) has a graph as shown then continuity at x = 0 is meaninglessSimilarly if f (x) has a graph as shown then continuity at x = 0 is meaningless -1 1 Also continuity at x = a ⇒ existence of limit at x = a but existence of limit atx a= ⇒/ continuity at x =a. Note: It should be remembered that all polynomial functions, trigonometric function, exponential and logarithmic functions are continuous in their domain. 67
- 68. Discontinuity of a Function A function f which is not continuous at a point ‘a’ is said to be discontinuous at the point ‘a’. ‘a’ is called a point of discontinuous of f or f is said to have a discontinuity at ‘a’. A function which is discontinuous even at a single point of an interval is said to be discontinuous in the interval A function f can be discontinuous at a point x = a because of any one of the following reasons: i) f is not defined at ‘a’ ii) lim ( ) x a f x → does not exist i.e., lim ( ) lim ( ) x a x a f x f x → − → + ≠ iii) lim ( ) ( )x a f x and f a→ both exist but are not equal 68
- 69. Types of Discontinuities DISCONTINUITY FIRST KIND SECOND KIND INFINITE OSCILLATORY REMOVABLE NON-REMOVABLE 69
- 70. Type-1 (Removable discontinuity) Here lim ( ) x a f x → necessarily exists, but is either not equal to f (a) or f (a) is not defined. In this case, therefore it is possible to redefine the function in such a manner thatredefine the function in such a manner that lim ( ) ( ) x a f x f a → = and thus making the function continuous. These discontinuities can be further classified as 70
- 71. Missing point discontinuity a) 2 ( 1)(9 ) ( ) 1 x x f x x − − = − at x =1 9 8 b) 2 4 ( ) 2 x f x x − = − at x=2 4 0 1 3-3 0 2 c) sin ( ) x f x x = 1 0π− 2 π − 2 π π 2 π 71
- 72. Isolated point discontinuity a) f (x) = [x]+[–x] 0 1 if x I if x I ∈ = − ∉ 1 y -2 2-1 b) f(x)=sgn(cos2x–2sinx+3) = 2(2 + sin x)(1 – sinx) 0 2 2 1 2 if x n if x n π π π π = + = + ≠ + 1 x -2 1 1 2 2-1 -2 -1 O 1 2 2 if x n π π+ ≠ + has an isolated point at x = 0 discontinuity as x=2n + 2 π π x y 1 3 2 π O 2 π 5 2 π 72
- 73. Type-2 (Non-Removable discontinuity) Here 1 lim ( ) x f x → does not exists and therefore it is not possible to redefine the function in any manner to make it continuous. Such discontinuities can be further classified into three types. (a) Finite type (both limits finite and unequal) (b) Infinite type (at least one of the two limit are infinity) (c) Oscillatory (limits oscillate between two finite quantities) 73
- 74. Finite type Example: 1 1 0 1 lim tan x x − → (0 ) 2 (0 ) 2 f f π π + − = = − ; jump π= Example: 2 ; jump 2= | sin | lim x (0 ) 1f + = Example: 2 ; jump 2= 0 lim x x→ (0 ) 1f − = − 1 ; jump 2 = 2 [ ] lim x x x→ (2 ) 1 1 (2 ) 2 f f + − = = Example: 3 Note : In this case non negative difference between the two limits is called the Jump of discontinuity. A function having a finite number of jumps in a given interval I is called a PIECE WISE CONTINUOUS or SECTIONALLY CONTINUOUS function in this interval. 74
- 75. Infinite type Example: 1 ( ) 1 1 x f x at x x = = − (1 ) (1 ) f f + − = −∞ = +∞ 0 2 f π + = Example: 2 Example: 3 2 1 ( ) 0f x at x x = = (0 ) (0 ) f f + − =∞ =∞ tan ( ) 2 2 x f x at x π = = 0 2 2 f f π − = =∞ 75
- 76. Oscillatory Type Example: 1 1 ( ) sinf x x = at x = 0 Oscillates between – 1 & 1 Example: 2 1 ( ) cosf x x = at x = 0 Oscillates between – 1 & 1 Example: 3 oscillates between 0 & 1 at x = 0 1 ( ) 1 sin(ln| |) 3 f x x = + 76
- 77. In Brief Removable Discontinuity If lim ( ) x a f x → exists but is not equal f(a) , the f is said to have a removable discontinuity at ‘a’. This type of discontinuity can be removed by defining a new function g as ( ) ( ) lim ( ) x a f x if x a g x f x if x a → ≠= = The g is continuous at ‘a’ Note: If lim ( ) x a f x → does not exist, then the function cannot be made continuous, no matter how define f(a). 77
- 78. Discontinuity of First Kind (or Jump Discontinuity) If lim ( ) x a f x → − and lim ( ) x a f x → + both exist but are unequal, then f is said to have a discontinuity of first kind at ‘a’ or jump discontinuity at ‘a’discontinuity of first kind at ‘a’ or jump discontinuity at ‘a’ f is said to have a discontinuity of the first kind from the right at ‘a’ if lim ( ) x a f x → + exists but is not equal to f(a) 78
- 79. Discontinuity of Second kind If neither lim ( ) x a f x → − nor lim ( ) x a f x → + exist, then f is said to have a discontinuity of second kind at ‘a’. f is said to have a discontinuity of the second kind from the left at ‘a’ if lim ( ) x a f x → − does not x a→ − exist. f is said to have a discontinuity of the second kind from the right at ‘a’ if lim ( ) x a f x → + does not exist 79
- 80. Mixed Discontinuity If a function f has a discontinuity of the second kind on one side of ‘a’ and on the other side, a discontinuity of the first kind or may be continuous, then f is said to have a mixed discontinuity at ‘a’ Thus f has a mixed discontinuity at ‘a’ if eitherThus f has a mixed discontinuity at ‘a’ if either (i) lim ( ) x a f x → − does not exist and lim ( ) x a f x → + exists, however lim ( ) x a f x → + may or may not equal f(a) (Or) (ii) lim ( ) x a f x → + does not exist and lim ( ) x a f x → − exists, however lim ( ) x a f x → − may or may not equal f(a). 80
- 81. Piecewise continuous Function A function :f A R→ is said to be piecewise continuous on A if A can be divided into a finite number of parts so that f is continuous on each part. Clearly, in such a case, f has a finite number of discontinuities and the set A isClearly, in such a case, f has a finite number of discontinuities and the set A is divided at the points of discontinuities . For example, consider : (0,5)f R→ defined by f(x)=[x], then f is discontinuous at 1, 2, 3 and 4. If the interval (0,5) is divided at 1, 2, 3 and 4, then f is continuous in (0,1), (1,2), (2, 3), (3, 4) and (4, 5) ∴ f is piecewise continuous 81
- 82. Property-1: Let f and g be two functions of x defined in some nbd of c .If f and g are both continuous at c . Then Elementary Properties of Continuous Function (i) f + g, f-g and f.g are all continuous at c (ii) f/g is also continuous at c provided that ( ) 0g c ≠ (iii) f or g is continuous at c. 82
- 83. Property-2:- Let f be defined on a set D with range R .Let g(u) be defined on a set D′ which contains R . If x D∈ , then ( )f x R∈ and consequently f(x) = u∈ D′ so that g{f(x)}is defined. The resulting function F, defined by F(x)=g{f(x)}, is called composition of g with f. we say that the composite function F is defined on D.with f. we say that the composite function F is defined on D. Now let f and g satisfy the conditions of the above paragraph. If f and g are both continuous at a point c D∈ , then the composite function F=g(f) is also continuous at c. Note: Continuous function of a continuous function is continuous. Property-3:- If a function f has its values 0≥ ,and if f is continuous at x=c , then f is also continuous at c. 83
- 84. Neighbourhood Property: If f is continuous at a certain point c and if f(c) ≠ 0, then there exists a certainIf f is continuous at a certain point c and if f(c) ≠ 0, then there exists a certain nbd of c such that ∀ x in that nbd of f(x) has the same sign as that of f(c). 84
- 85. Deeper Properties Of Continuous Functions Let f be a real-valued continuous function in a closed interaval [a,b]. Suppose that f(a)and f(b) are of opposite signs i.e f(a).f(b) < 0. Then there exist at least one point c where a < c < b such that f(c) = 0. BOLZANO’S THEOREM ON CONTINUITY: a b f(a)>0 f(b)<0 α 85
- 86. Let f be a real-valued continuous function in a closed interaval [a,b].Suppose that f(a) ≠ f(b).Then f assumes every value between f(a) and f(b) at least once. INTERMEDIATEVALUE THEOREM:- 86
- 87. FIXED-POINT THEOREM:- If f is continuous on[a,b] and f(x) ∈[a,b] for every x∈[a,b] (That is :[ , ] [ , ]f a b a b→ ) then f has a fixed point i.e ∃ c ∈[a,b] whose f(c)=c Theorem:-Theorem:- Let f be a real –valued continuous function in [a,b] .Suppose that f(a) ≠ f(b).Let µbe a number between f(a) and f(b) .Moreover,assume that f is strictly monotone in[a,b].Then ∃ exactly one value of x = c where a < c < b such that f(c)= µ. 87
- 88. CONVESE OF INTERMEDIATE VALUE THEOREM: If in [a,b], f assumes at least once every value between f(a) and f(b) with f(a) ≠ f(b), then’f’ need not necessarily be continuous in [a,b]. Example: Let f be function defined by 0, 0 1 1 0 2 2 1 1 ( ) 2 2 3 1 1 2 2 1, 1 when x x when x f x when x x when x when x = − < < = = − < < = 88
- 89. 1 C (1,1) Clearly , there are discontinuous at x=0,1/2,1.But every value between 0 and 1 is assumed by f.Take a line parallel to X-axis between y=0 and y=1[e.g.,a dotted line shown in the figure , y = d(0 1d≤ ≤ ). ∃a corresponding x=c for f(c)=d].Thus f assumes every value between f(0) and f(1) but f is not continuous in0 1x≤ ≤ . 1 d C D CB 1 (1/2,1/2)1/2 (1,1) (0,0) 1/2 A O 89
- 90. BOUNDEDNESS PROPERTY OF CONTINUOUS FUNCTION:- If a real valued function f is continuous on a closed interval I=[a,b]then it is bounded there. NOTE:-A function which is continuos only in (a,b) or [a,b) or (a,b] may not be bounded there. Example:- Consider 1 ( ) (0,1)f x x x = ∀ ∈ Which is continuous in (0,1) also as 0 ( )x f x+ → ⇒ → +∞Which is continuous in (0,1) also as 0 ( )x f x+ → ⇒ → +∞ I,e it is not bounded NOTE:- CONVERSE OF THE ABOVE THEOREM NEED NOT BE TRUE. i.e a function which is bounded in an interval need not be continuous throughout the interval . Example:-Consider ;0 1 ( ) 1 ; 0,1 2 x x f x when x < < = = Clearly f is bounded in between 0 and 1 but not continuous in 0 1x≤ ≤ . 90
- 91. ATTAINMENT OF BOUNDS OF CONTINUOUS FUNCTION: Property-1: If f be continuous function on a closed interval, then it assumes its least upper bound(l.u.b) and greatest lower bound (g.l.b) in that interval. Remember:Remember: In a bounded set of real numbers if the lub and glb are actually members of the set , then they are respectively called greatest and least value of the set .So A function f is continuous in [a,b] then it has greatest and least values in [a,b]. If f is continuous in (a,b) then the above statement need not be true. 91
- 92. Example:- Consider 1 ;0 1 ( ) 1 ; 0 2 x x f x when x − < ≤ = = 2 At x=0 , there is a point of discontinuity i.e., f is not continuous in [0,1] and f has no greatest value in [0,1]. Property-2: If a function f is continuous in [a,b] then f assumes all values between its exact upper and lower bounds. 92
- 93. CONTINUITY OF INVERSE FUNCTION Property-1: Let f be a strictly increasing function on [a,b].Then (i) The inverse function f-1 exists and is strictly monotone increasing in its domain of definition (ii) If , further , f is continuous in [a,b] then f-1 is also continuous on [ ],α β where(ii) If , further , f is continuous in [a,b] then f is also continuous on [ ],α β where ( )f aα = and ( )f bβ = Let f be a strictly decreasing function on [a,b].Then (iii) The inverse function f-1 exists and is strictly monotone decreasing in its domain of definition (iv) If , further , f is continuous in [a,b] then f-1 is also continuous on [ ],β α where ( )f aα = and ( )f bβ = Remember:- If f is strictly increasing then –f is strictly decreasing and vice – versa 93
- 94. CONTINUITYAND MONOTONICITY If f be a continuous in [a,b] and assumes each value between f(a) and f(b) just once then it is strictly monotonic in the same interval Remember: Let f be monotonic increasing in [a,b] .Then if c be s.t a<c<b then f(c-0) and f(c+0) both exist and sup ( 0) a x c f f c < < = − and inf ( 0) c x b f f c < < = + also ( 0) ( ) ( 0)f c f c f c− ≤ ≤ + 94
- 95. Conceptual Examples (1) State the number of point of discontinuities and discuss the nature of discontinuity for the function 1 ( ) ln| | f x x = and also sketch its graph. 1 0, 1 ln ( ) 1 0, 1 ln( ) if x x x f x if x x x > ≠ = < ≠ − − function is obviously discontinuous at x = 0, 1, –1. as it is not defined. 95
- 96. 0 0 lim ( ) 0 lim ( ) 0 x x f x f x + − → → = = Limit exists at x = 0. Hence removable discontinuity at x = 0. (Missing point) 1 lim ( ) x f x+ → = ∞ Limit DNE. Hence non removable discontinuity (infinite type) at1 1 lim ( ) x x f x− → → = −∞ Limit DNE. Hence non removable discontinuity (infinite type) at x = 1 1 1 lim ( ) lim ( ) x x f x f x + − →− →− = −∞ = ∞ Limit DNE. Hence non removable discontinuity (infinite type) at x = 0 96
- 97. Note that f (x) is even ⇒ symmetric about y axis. The graph of f (x) is as follows. 97
- 98. (2) If f(x+y) = f(x) . f(y) for all x & y & f(x) = 1 + g(x). G(x) where 0 lim ( ) 0 x g x → = & 0 lim ( ) x G x → exist. Prove that f(x) is continuous for all x. Sol. Let x a R= ∈ 0 0 0 lim ( ) lim ( )· ( ) lim ( )[1 ( )· ( )] h h h f a h f a f h f a g h G h → → → + = = + 0 0 ( ) 1 lim ( )·lim ( ) ( ) h h f a g h G h f a → → = + = 98
- 99. (3) Show that the function f (x) = (x – a)2 (x – b)2 + x takes the value 2 a b+ for some value of x ∈ [a, b] Sol. f (a) = a ; f (b) = b ; Also find f is continuous in [a, b] and 2 a b+ ∈ [a, b] Hence using intermediate value theroem + ∃ some c ∈ [a, b] such that ( ) 2 a b f c + = 99
- 100. (4) Prove that there exist a number x such that 2005 2 1 x 2005 1 sin x + = + Sol Let 2005 2 2 f (x) = x + (1 + sin x)− ∴ f is continuous and f (0) = 1 < 2005 and f (2) > 22005 ∴ f is continuous and f (0) = 1 < 2005 and f (2) > 2 100
- 101. (5) Let f be a continuous function defined onto on [0, 1] with range [0, 1]. Show that is some 'c' in [0, 1] such that f (c) = 1 – c. Sol. Consider g (x) = f (x) – 1 + x g (0) = f (0) – 1 ≤ 0 (as f (0) ≤ 1)g (0) = f (0) – 1 ≤ 0 (as f (0) ≤ 1) g (1) = f (1) ≥ 0 (as f (1) ≥ 1) Hence g (0) and g (1) and of opposite signs hence ∃ at least one c ∈ (0, 1) such that g (c) = 0 ∴ g (c) = f (c) – 1 + c = 0 ; f (c) = 1 – c 101
- 102. (6) Let f be continuous on the interval [0, 1] to R such that f (0) = f (1). Prove that there exists a point c in 1 0, 2 such that 1 ( ) 2 f c f c = + Sol. Consider a continuous function 1 g (x) = f f (x) 2 x + ( g is continuous 1 0, 2 x ∀ ∈ )g (x) = f f (x) 2 x + ( g is continuous 0, 2 x∀ ∈ ) now 1 1 g (0) = f f (0) = f f (1) 2 2 [as f (0) = f (1)] 1 1 1 (1) (1) 2 2 2 g f f f f = − = − since g is continuous and g (0) and g(1/2) are of opposite signs hence the equation g (x) = 0 must have at least one root in 1 0, 2 ∴for some 1 1 0, , ( ) 0 ( ) 2 2 c g c f c f c ∈ = ⇒ + = 102
- 103. (7) Let f : [0, 2] → R be continuous and f (0) = f (2). Prove that there exists x1 and x2 in [0, 2] such that x2 – x1 = 1 and f (x2) = f (x1) Sol. Consider continuous function g as g (x) = f (x + 1) – f (x) now, g (0) = f (1) – f (0) = f (1) – f (2) ....(1)now, g (0) = f (1) – f (0) = f (1) – f (2) ....(1) g (1) = f (2) – f (1) = f (2) – f (1) ....(2) hence g (0) and g (1) are of opposite signs, hence ∃ some c ∈ (0, 1) where g (c) = 0 i.e. f (c + 1) = f (c) [c + 1 ∈ (1, 2) as c ∈ (0, 1) ] put c = x1 ; c + 1 = x2 ∴ f (x2) = f (x1) where x2 – x1 = 1 obviously x1, x2 ∈ [0, 2] 103
- 104. (8) 2 sin ( ) 2 x x f x x = + & g (x) = | x | are continuous at x = 0, hence the composite2 2x + 2 sin (gof) (x) 2 x x x = + will also be continuous at x = 0. 104
- 105. (9) Dirichlet’s Function Let 1, ( ) 0, c if x Q f x if x Q ∈ = ∈ Prove that f is NOT continuous at any point of R . Sol: Let { }nx be a sequence of rational numbers such that nx a→ and nx a≠ , n=1,2,3…..;Let { }ny be a sequence of irrational numbers such that ny a→ and{ }n n ny a≠ , n=1,2,3…..;.Clearly ( ) 1, ( ) 0n nf x f y= = for n=1,2,3…… Therefore ( ) 1nf x → and ( ) 0nf y → and lim ( ) x a f x → does not exist. Since ‘a’ is arbitrary, limit does not exist for every real number. Hence it is not continuous every where. (10) Let 0, ( ) 1, c if x Q f x if x Q ∈ = ∈ is discontinuous every where. Sol: Similar to above Solution 105
- 106. (11) Thomac Function Let f be defined on R+ by 0, 1 ,( ) int , 1 if xisirrational p if x where p and q aref x q q are ve egers and hcf of p qis == + int , 1are ve egers and hcf of p qis+ Prove that f is continuous at every irrational number in R+ and discontinuous at every rational number in R+ Sol: Case 1: Suppose ‘a’ is any positive rational number. Let { }nx be a sequence of irrational numbers in R+ that converges to a. Then lim ( ) 0n n f x →∞ = while f(a) = a positive number of the form 1 q where q is a positive integer. Hence f is discontinuous at ‘a’ 106
- 107. Case 2: Suppose ‘b’ is an irrational number in R+ . Let ε >0 .Then by the Archimedean property , ∃ a positive integer n0 such that 0 1n ε > .There are only a finite number of rational numbers with denominator less than n0 in the open interval (b-1,b+1).interval (b-1,b+1). Hence δ > 0 can be chosen so small that the nbd ( ),b b− δ + δ contains no rational numbers with denominators less than n0. It then follows that for x b− < δ and x ∈R+ we have 0 1 ( ) ( ) ( )f x f b f x n − = ≤ < ε . Thus f is continuous at the irrational number b. 107
- 108. (12). Let , ( ) 1 , c x if x Q f x x if x Q ∈ = − ∈ Prove that f is NOT continuous at any point of R except at x = 1 2 Sol: At x = 1 2 , 1 1 2 2 f = 1 1 ( ) , 2 2 f x f x if x Q − = − ∈ 1 (1 ) 1 , 2 1 , 2 C C x if x Q x evenif x Q = − − − ∈ = − ∈ Therefore given anyε >0, we can make 1 1 ( ) ( ) 2 2 C f x f x Q Q x − < ε∀ ∈ ∪ − < δ Where ε =δ i.e f is continuous at x= 1 2 . 108
- 109. Differentiation && Differentiability
- 110. Concepts 110 Points to remember
- 111. Derivative of a Function at a Point Let :[ , ]f a b R→ be a function and ( , )c a b∈ , then f is said to be derivable (or differentiable) at c, if ( ) ( ) ( ) ( ) lim lim f x f c f c h c c or x c h − + − − exists. The limit, indifferentiable) at c, if 0 lim lim x c h or x c h→ →− exists. The limit, in case it exists is called the derivative or the differential co-efficient of the function at x =c and is denoted by ( )f c′ 111
- 112. Left-Hand Derivative(LHD) Let :[ , ]f a b R→ be a function and ( , )c a b∈ If ( ) ( ) lim x c f x f c x c→ − − − or 0 ( ) ( ) lim h f c h f c h→ + − − − exists, the this limit is called by the left- hand derivative of f at c and is denoted by ( 0)f c′ − or ( )f c′ − or ( )Lf c′hand derivative of f at c and is denoted by ( 0)f c′ − or ( )f c′ − or ( )Lf c′ Right-Hand Derivative(RHD) Let :[ , ]f a b R→ be a function and ( , )c a b∈ If ( ) ( ) lim x c f x f c x c→ + − − or 0 ( ) ( ) lim h f c h f c h→ + − exists, then this limit is called the right- hand derivative of f at c and is denoted by ( 0)f c′ + or ( )f c′ + or ( )Rf c′ . 112
- 113. Existence of Derivative at a point A function f is said to be differentiable at x =c iff i) ( )Lf c′ exists (which is fixed finite quantity) ii) ( )Rf c′ exists (which is fixed finite quantity) iii) ( ) ( )Lf c Rf c′ ′= If any one of the above condition fails to satisfy by the function f(x) then f is said to be not derivable at x =c. The derivative ( )f c′ exists iff ( ) ( )Lf c Rf c′ ′= . Note: When we say that ( ) ( ) lim x c f x f c x c→ − − exists, we mean that it is a real number i.e, finite 113
- 114. Derivability in an interval A function :[ , ]f a b R→ is said to be derivable in the open interval (a, b) if ( )f c′ exists for each c such that a c b< < A function :[ , ]f a b R→ is said to be derivable in the closed interval [a,b] ifA function :[ , ]f a b R→ is said to be derivable in the closed interval [a,b] if (i) ( )f c′ exists for each ( , )c a b∈ (ii) ( )Rf a′ exists, and (iii) ( )Lf b′ exists A function :f I R→ is said to be derivable on I if f is derivable at every point of I. 114
- 115. Derivative of a Function (w.r.t x) Let f be a function whose domain is an interval I. Let I1 be the set of all those points x of I at which ( )f x′ exists. Clearly 1I I⊂ . If 1I ≠ φ , then the function f ′ with domain I1 is called the derivative of f i.e., 0 ( ) ( ) lim h f x h f x h→ + − exists fx D∀ ∈ then f is said to be differentiable w.r.t x and is denoted by ( ) ( ) d dy f x or or f x dx dx ′ . This definition is called ab-intio method or First principles of Differentiation. And hear d dx differential co-efficient of y=f(x) w.r.t x. 115
- 116. Interpretation of Derivative Geometrical Interpretation : Slope of the tangent drawn to the curve at x = a if it exists Physical Interpretation: Instantaneous rate of change of functionInstantaneous rate of change of function Remember : If graph of the function contains any sharp edges, corners, gaps or peaks then at that points the function is not differentiable. 116
- 117. Properties, Theorems In Differentiation Theorem-1:- If f has a finite derivative at x = c, then f is continuous at x=c. i.e., derivability ensures continuity Remember:- 1.Contrapositive of the above statement is “ If a function f is not continuous at a point x=c , then the function f can not be derivable there.” 2. Every continuous function need not be derivable 117
- 118. DARBOUX THEOREM: Assume that f is defined on the closed interval [a,b] and that f has finite derivative at each interior point .Assume also that f has a RHD at x=a and LHD at x=b where ( ) ( ) 0f a f b′ ′ < .Then there exist at least one point c in [a,b] such that ( ) 0f c′ = .( ) ( ) 0f a f b′ ′ < .Then there exist at least one point c in [a,b] such that ( ) 0f c′ = . Note:- If ( ) 0f a′ < and ( ) 0f b′ > then with similar arguments it can be shown that there exist an interior point [ , ]d a b∈ where ( ) 0f d′ = .In this case d is the point where f attains the Infimum. 118
- 119. INTERMEDIATE VALUE PROPERTY FOR DERIVATIVES Let f be a deriavable function in [a,b] and ( ) ( )f a f b′ ′≠ .Let γ be any number between ( )f a′ and ( )f b′ .Then ∃atleast one point [ , ]c a b∈ ∋ ( )f c′ = γ Observation: Let f be a real-valued continuous function in a closed interaval [a,b].Suppose that f(a) ≠ f(b).Then f assumes every value between f(a) and f(b) at least once. This is what is know as the intermediate value property of continuous function of [a,b]. This property does not characterize continuous function, because we have just proved that the property is also shared by the class of derivatives ( )f x′ , whether continuous or not. 119
- 120. Formulae & Shortcuts of Differentiation If U = U(x), V = V(x) then 1. ( ) d dU dV U V dx dx dx ± = ± 2. ( ) d dU dV aU bV a b dx dx dx + = ⋅ + ⋅ (where a, b are any constants)2. ( )aU bV a b dx dx dx + = ⋅ + ⋅ (where a, b are any constants) 3. ( ) d dV dU UV U V dx dx dx = ⋅ + ⋅ (Product rule) 4. 2 dU dV V U d U dx dx dx V V ⋅ − ⋅ = (Quotient Rule) 120
- 121. 5. 1dy dxdx dy = 6. ( ) ( ) 1 ( ) ( ) ( ) n nd f x n f x f x dx − ′= ⋅ ⋅ 7. ( ) ( ) ( )f x f xd e e f x′= ⋅7. ( ) ( ) ( )f x f x e e f x dx ′= ⋅ 8. ( ) ( ) log ( )f x f xd a a a f x dx ′= ⋅ ⋅ 9. 1 log ( ) ( ) | ( ) | d f x f x dx f x ′= 10. 1 log | | d x dx x = 11. | | , ( 0) | | d x x x dx x = ≠ 121
- 122. 12. sin ( ) cos ( ) d f x f x dx = 13. cos ( ) sin ( ) d f x f x dx =− 14. tan ( ) sec² ( ) d f x f x=14. tan ( ) sec² ( ) d f x f x dx = 15. cot ( ) cosec² ( ) d f x f x dx =− 16. sec ( ) sec ( ) tan ( ) d f x f x f x dx = ⋅ 17. cosec ( ) cosec ( ) cot ( ) d f x f x f x dx =− ⋅ 122
- 123. 18. ( ) 1 2 1 sin ( ) ( ) 1 ( ) d f x f x dx f x − ′= ⋅ − 19. ( ) 1 2 1 cos ( ) ( ) 1 ( ) d f x f x dx f x − ′=− ⋅ − 20. 1 1 tan ( ) ( ) d f x f x− ′= ⋅20. ( ) 1 2 1 tan ( ) ( ) 1 ( ) d f x f x dx f x − ′= ⋅ + 21. ( ) 1 2 1 cot ( ) ( ) 1 ( ) d f x f x dx f x − ′=− ⋅ + 22. ( ) 1 2 1 sec ( ) ( ) | ( ) | ( ) 1 d f x f x dx f x f x − ′= ⋅ − 23. ( ) 1 2 1 cosec ( ) ( ) | ( ) | ( ) 1 d f x f x dx f x f x − ′=− ⋅ − 123
- 124. 24. If y=(f(x))g(x) , then ( ) 1 g(x) g(x) 1dy d f (x) f (x) (f (x)) g(x). g (x)log(f(x)) dx dx f(x) = = + 25. If y=(f(x))f(x) , then dx dy =(f(x))f(x) (1+log(f(x))f1 (x)25. If y=(f(x)) , then dx =(f(x)) (1+log(f(x))f (x) 26. If y=(f(x))y , then )log1)(( )( ))(log(1)(( )( 1212 yxf xfy xfyxf xfy dx dy − = − = 27. If yxfxfxfxfy +=−∞−−−−+++= )()()()( , then 12 )(1 − = y xf dx dy 124
- 125. 28. =∞−−−−−= dx dy then,)()()( xfxfxfy f1 (x) 29. If f(x,y)=kg(x,y), where f(x,y) and g(x,y) are the homogenous functions of same order, then ydy = x y dx dy = 30. If f(x,y)= C, then f dy x dx f y ∂ ∂ = − ∂ ∂ 31. If f(x+y)=f(x)f(y) ∀ x,y ∈ R and f(x)≠0, f(a)=k, f1 (0) exists, then f1 (a)=kf1 (0) 125
- 126. 32. If y=Tan-1 ( ) ( ) ( ) ( ) = − + dx dy then, )()( )()( xfaSinxfbCos xfbSinxfaCos f1 (x) 33. If y=Tan-1 ( ) ( ) ( ) ( ) = + − dx dy then, )()( )()( xfaSinxfbCos xfbSinxfaCos -f1 (x) R 34. If y= 3 2 1 R R R , where R1,R2,R3 are the rows of the determinant, whose the elements are the polynomials in ‘x’, then 1 3 2 1 3 1 2 1 3 2 1 1 R R R R R R R R R dx dy ++= and det(A)=det(AT ) ⇒ we can apply the same method in column wise also 126
- 127. 35. If ( )nnxx yxkyx −=−+− 22 11 , then n nn x y y x dx dy 2 21 1 1 − − = − 36. If ( )nnnn yxkyx +=−−− 22 11 , then n nn x y y x dx dy 2 21 1 1 − − = − ∂ − f 37. If aSin( ) ( ) ∂ ∂ ∂ ∂ − ==+ y f x f cyxfbCosyxf dx dy then,),(),( 38. If y=Tan-1 )(2dx dy then, 2 22 bCosxa bax Tan ba ba + − = + − 39. If bCosxa ba ba bay Tan + − = + − = 22 dx dy then, 2 x Tan 2 127
- 128. 40. If y=Cos-1 bCosxa ba ba bCosxa baCosx + − => + + 22 dx dy then),( 41. If y=Cos(mCos-1 x) or y=Cos(mSin-1 x) or y=Sin(mSin-1 x) or y=Sin(mCos-1 x), then (1-x2 )y2-xy1+m2 y=0 42. If y=Cos(mCos-1 x) (or) y=Cos(mSin-1 x) (or) y=Sin(mCos-1 x) (or) y=Sin(mSin-1 x),42. If y=Cos(mCos-1 x) (or) y=Cos(mSin-1 x) (or) y=Sin(mCos-1 x) (or) y=Sin(mSin-1 x), then (1-x2 )y2-xy1+m2 y=0 43. If y=k1eax , then (D-a)y=0, where Dy= dx dy 44. If y=(k1+k2x)eax , then (D-a)2 y=0, where D2 y= 2 2 dx yd 45. If y=k1eax +k2ebx , then (D-a)(D-b)y=0 128
- 129. 46. If y=eax Cos(bx+c) (or) y=eax Sin(bx+c), then ((D-a)2 +b2 )y=a 47. If ax2 +2hxy+by2 =c, then y2= 3 2 )( )( byhx abhc + − 48. If y=k1Sin(ax+b)+k2Cos(ax+b), then (D2 -a2 )y=0 49. If y=k1eax Sin(bx+c)+k2eax Cos(bx+c), then ((D-a)2 +b2 )y=0 50. If y=xn logx, then yn= 1 1 n! logx 1 2 n + + +−−−−−+ 51. If y=xn-1 logx, then yn= n! 1 x − 129
- 130. Good LuckGood Luck

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