Chapter 2.2 Functions, Limits, Continuity and Differentiability PDF
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This document describes functions, limits, continuity, and differentiability, providing definitions, theorems, and methods for evaluating limits. It includes various examples and important expansions for different functions like logarithmic, exponential, trigonometric. This document is suitable for undergraduate level mathematics.
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Functions, Limits, Continuity and Differentiability 60 69 2.2.1 Limit of a Function. Let y f (x ) be a function of x. If at x a, f ( x ) takes indeterminate form, then we consider the values of E3 the function which are very near to ‘a’. If these values tend to a definite unique number as x tend...
Functions, Limits, Continuity and Differentiability 60 69 2.2.1 Limit of a Function. Let y f (x ) be a function of x. If at x a, f ( x ) takes indeterminate form, then we consider the values of E3 the function which are very near to ‘a’. If these values tend to a definite unique number as x tends to ‘a’, then the unique number so obtained is called the limit of f (x ) at x a and we write it as lim f (x ). x a a U x ID (1) Meaning of ‘x a’: Let x be a variable and a be the constant. If x assumes values nearer and nearer to ‘a’ then we say ’x tends to a’ and we write ' x a'. It should be noted that as x a , we have x a. By ' x tends to a' we mean that (i) x a (ii) x assumes values nearer and nearer to ‘a’ and (iii) We are not specifying any manner in which x should approach to ‘a’. x may approach to a from left or right as shown in figure. a x D YG (2) Left hand and right hand limit : Consider the values of the functions at the points which are very near to a on the left of a. If these values tend to a definite unique number as x tends to a, then the unique number so obtained is called left-hand limit of f (x ) at x = a and symbolically we write it as f (a 0) lim f (x ) lim f (a h) x a h 0 Similarly we can define right-hand limit of f (x ) at x a which is expressed as f (a 0) lim f (x ) lim f (a h). h 0 x a U (3) Method for finding L.H.L. and R.H.L. (i) For finding right hand limit (R.H.L.) of the function, we write x + h in place of x, while for left hand limit (L.H.L.) we write x – h in place of x. ST (ii) Then we replace x by ‘a’ in the function so obtained. (iii) Lastly we find limit h 0. (4) Existence of limit : lim f (x ) exists when, x a (i) lim f (x ) and lim f (x ) exist i.e. L.H.L. and R.H.L. both exists. x a x a (ii) lim f (x ) lim f (x ) i.e. L.H.L. = R.H.L. x a Note : x a If a function f (x ) takes the form 0 or at x a , then we say that f (x ) is indeterminate or 0 meaningless at x a. Other indeterminate forms are , , 0 , 1 ,0 0 , 0 In short, we write L.H.L. for left hand limit and R.H.L. for right hand limit. 70 Functions, Limits, Continuity and Differentiability It is not necessary that if the value of a function at some point exists then its limit at that point must exist. (5) Sandwich theorem : If f (x ) , g(x ) and h(x ) are any three functions such that, f (x ) g(x ) h(x ) x neighborhood of x a and lim f (x ) lim h(x ) l (say) , then lim g(x ) l. This theorem is normally applied when x a x a x a the lim g(x ) can't be obtained by using conventional methods as function f (x ) and h(x ) can be easily found. x , when x 1 If f (x ) 2 , then lim f (x ) x 1 x , when x 1 (a) x 2 Solution: (d) [MP PET 1987] (b) x (c) – 1 (d) 1 E3 Example: 1 60 x a To find L.H.L. at x 1. i.e., lim f (x ) lim f (1 h) = lim (1 h)2 = lim (1 h 2 2 h) = 1 i.e., lim f (x ) 1 x 1 Now find R.H.L. at x = 1 i.e., lim f (x ) lim f (1 h) = 1 i.e., x 1 ….(i) x 1 h 0 h0 h 0 lim f (x ) 1 x 1 …..(ii) ID h 0 From (i) and (ii), L.H.L. = R.H.L. lim f (x ) 1. x 1 lim x 2 | x 2| x 2 U Example: 2 (a) 1 L.H.L.= lim x 2 (c) Does not exist and, R.H.L.= lim x 2 x 2 lim f (x ) 0 (b) U x 3 lim f (x ) 0 (c) x 3 lim f (x ) lim f (x ) x 3 x 3 (d) None of these 2 1 53 3 x , if 0 x 1 Let the function f be defined by the equation f (x ) , then 5 3 x , if 1 x 2 (a) Solution: (d) | x 2| does not exist. x 2 x 3 lim f (x ) 5 3 2 and lim f (x ) x 3 ST Example: 4 …..(ii) 2 , when x 3 If f (x ) 5 x , then 5 x , when x 3 (a) Solution: (c) …..(i) h | x 2| | 2 h 2| lim = lim 1 h0 h h 0 2 h 2 x 2 From (i) and (ii) L.H.L. R.H.L. i.e. lim Example: 3 (d) None of these h | x 2| | 2 h 2| 1 lim = lim h 0 h h 0 2 h 2 x 2 D YG Solution: (c) (b) –1 lim f (x ) f (1) (b) x 1 lim f ( x ) 3 (c) x 1 lim f ( x ) 2 x 1 [SCRA 1996] (d) lim f (x ) does not exist x 1 L.H.L. lim f (x ) lim f (1 h) lim 3(1 h) lim (3 3 h) 3 3.0 3 x 10 h0 h0 h 0 R.H.L. lim f (x ) lim f (1 h) lim [5 3(1 h)] lim (2 3 h) 2 3. 0 2 x 1 0 h 0 h 0 h 0 Hence lim f (x ) does not exists. x 1 Example: 5 lim x 0 | x| x (a) 1 [Roorkee 1982; UPSEAT 2001] (b) –1 (c) 0 (d) Does not exist Functions, Limits, Continuity and Differentiability 71 | x| | x| lim 1 and lim 1 , hence limit does not exists. x 0 x x 0 x Solution: (d) 2.2.2 Fundamental Theorems on Limits. The following theorems are very useful for evaluation of limits if lim f (x ) l and lim g(x ) m (l and m are x 0 (1) lim ( f (x ) g(x )) l m (Sum rule) (2) lim ( f (x ) g(x )) l m x a x a (4) lim k f (x ) k.l (3) lim ( f (x ).g(x )) l.m (Product rule) x a x a f (x ) l ,m 0 g(x ) m (Quotient rule) (6) If lim f (x ) or , then lim x a x a 1 0 f (x ) x a x a ID x a lim g( x ) (9) lim [ f (x )]g( x ) {lim f (x )} x a x a (Constant multiple rule) (8) If f (x ) g(x ) for all x, then lim f (x ) lim g(x ) (7) lim log{ f (x )} log {lim f (x )} x a (Difference rule) E3 (5) lim x a 60 x 0 real numbers) then x a x a U (10) If p and q are integers, then lim ( f (x )) p / q l p / q , provided (l) p / q is a real number. (11) If lim f (g(x )) f (lim g(x )) f (m ) provided ‘f’ is continuous at g(x ) m. e.g. lim ln[ f (x )] ln(l), only if x a x a D YG l 0. x a 2.2.3 Some Important Expansions. In finding limits, use of expansions of following functions are useful : n(n 1) 2 (1) (1 x ) 1 nx x ..... 2! n x2 x3 ..... 2! 3! (4) log(1 x ) x U (3) e x 1 x ST (5) log(1 x ) x (6) (1 1 x )x x2 x3 x4 ....... , 2 3 4 1 log(1 x ) ex x x2 1 2 3 e (2) a x 1 x log a (x log a) 2 ..... 2! x2 x3 x4 .....,| x | 1 2 3 4 where | x | 1 x 11 2 ....... e 1 x ....... 2 24 x2 x4 x6 ...... 2! 4 ! 6! (7) sin x x x3 x5 ....... 3! 5! (8) cos x 1 (9) tan x x x 3 2x 5 ..... 3 15 (10) sinh x x x3 x5 ..... 3! 5! (12) tanh x x x3 2 x 5 ..... 3 (11) cosh x 1 x2 x4 x6 ..... 2 ! 4! 6! 72 Functions, Limits, Continuity and Differentiability (13) sin 1 x x 1 2. (14) cos 1 x sin 1 x 2 x3 x5 x7 ..... 3 5 7 2.2.4 Methods of Evaluation of Limits. We shall divide the problems of evaluation of limits in five categories. 60 (15) tan 1 x x x3 x5 3 2.1 2. ..... 3! 5! (1) Algebraic limits : Let f (x ) be an algebraic function and ‘a’ be a real number. Then lim f (x ) is known x a E3 as an algebraic limit. (i) Direct substitution method : If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression. ID (ii) Factorisation method : In this method, numerator and denominator are factorised. The common factors are cancelled and the rest outputs the results. 1 1 , 2 3 etc.) on expressions in numerator or denominator or in both. After rationalisation the terms are factorised which on cancellation gives the result. U (iii) Rationalisation method : Rationalisation is followed when we have fractional powers (like (iv) Based on the form when x : In this case expression should be expressed as a function 1/x D YG and then after removing indeterminate form, (if it is there) replace 1 by 0. x Step I : Write down the expression in the form of rational function, i.e., f (x ) , if it is not so. g( x ) Step II : If k is the highest power of x in numerator and denominator both, then divide each term of numerator and denominator by x k. Step III : Use the result lim U x 1 xn 0 , where n > 0. ST Note : An important result : If m, n are positive integers and a0 , b 0 then lim x Example: 6 a 0 x m a1 x m 1 .... a m 1 x a m b 0 x n b 1 x n 1 ..... b n 1 x b n Example: 7 a0 b , if m n 0 0, if m n , if m n lim (3 x 2 4 x 5 ) x 1 (a) 12 Solution: (a) 0 are non-zero real numbers, (b) –1 (c) Does not exist (d) None of these lim (3 x 2 4 x 5) 3(1)2 4 (1) 5 12. x 1 The value of lim x 2 3x / 2 3 3x 9 is [MP PET 2000] Functions, Limits, Continuity and Differentiability 73 (a) 0 3x / 2 3 x 2 (3 ) (3) x/2 2 2 The value of lim x a (3 x / 2 3) = lim x 2 (3 x /2 Example: 9 lim x a x a lim 1 1 1 equals h x h x h 0 Solution: (d) Example: 10 n (x a) (x = lim x a [Rajasthan PET 1989, 92] lim h 0 1 h 1 x2 1 x2 The value of lim = lim (x n 1 x n 2 a .... an 1 ) = n. an 1. x a (c) x 3 lim x x 2 4 x (c) – 2 ax 2 bx c dx 2 ex f (a) ( x 3) 3 2 x 2 4 x x2 [MP PET 1999] (d) 0 [UPSEAT 1991] (c) 1 4 (d) None of these x 3 2 2 2 = b e (b) c f (c) a d (d) d a Here the expression assumes the form . We note that the highest power of x in both the numerator and denominator is 2. So we divide each terms in both the numerator and denominator by x 2. b c 2 x x a00 a. lim lim e f x dx 2 ex f x d 00 d d 2 x x ax 2 bx c Example: 13 1 x 2 4 x x 2 4 x 1 1 4x 1. = lim 2 x 3 ST x lim ( x 3) x 2 (2 x 6) U lim (d) 2 2 1 x 1 x 2 (1 x 2 ) (1 x 2 ) = lim 1. = 2 x 0 2 2 2 2 2 x 1 x 1 x 1 x 1 x (b) x 3 lim x 3 2 equals x 2 4 x x 3 1 U (b) –1 2 2 1 x 1 x lim x 0 x2 x 3 Solution: (c) ) 1 2x is x2 x 0 = lim Example: 12 x a .. a (x a) (d) 1 1 1 h 1 x (x h) 1 1 = 2. = hlim x h x = hlim 0 0 h ( x h ) x h ( x h ) x x (a) 1 Solution: (d) n 1 D YG Example: 11 n2 (b) (a) 1 Solution: (b) n 1 (c) na n [Rajasthan PET 1987] 1 2x (a) 1. 6 (b) nan 1 n x a = x n an is x a (a) 0 Solution: (b) 3)(3 x / 2 3) (d) ln 3 60 lim (c) E3 Example: 8 (b) 1 6 ID Solution: (c) 1 3 a lim x x x x is equal to x 74 Functions, Limits, Continuity and Differentiability Example: 15 Example: 16 lim x x x 1 (b) lim x x 1 lim x x lim x x 1 x 1 11 1 lim (1 x )1 / x x 1 (b) e lim (1 x )1 / x lim (1 x 1 x 1 The value of the limit of (a) 3 Example: 18 The value of the limit of 1 lim x 1 x x) x 3 x3 8 (x 2 4 ) 1 x 1 x 3 / 2 1 1. 2 (d) a 2, b 1 3 2 2 U ST 1 2 x lim x 0 1 x 1 x lim (a) a 2x 3x 3a x 2 x 2a 3 3 (d) 21 (c) 1 2 (b) 2 2x 4 x2 (c) 1 lim 1 x (d) 0 444 3. 22 [Rajasthan PET 1988] x 1 x 1x lim x 0 1 x 1 x 1 x 1 x x 1 x 1x lim x 0 1 x 1 x x a (d) None of these 2 ( x 2 x 4 )( x 2) x x 8 lim 2 lim lim x 2 x 2 x 2 x 4 ( x 2)( x 2) x lim is equal to x 0 1 x 1 x (a) (d) None of these as x tends to 2 is (b) 3 Example: 20 (c) Not defined x 3 x 2 18 as x tends to 3 is x 3 (b) 9 (c) 18 (a) 3 Solution: (c) 2 Let y lim Example: 19 x x x 18 lim (x 2 2 x 6) 9 6 6 = 21 x 3 x 3 3 Solution: (d) Solution: (a) x x x x (c) Not defined D YG Example: 17 1 x 1 / 2 lim x2 1 x 2 (1 a) x (a b) 1 b ax b 0 lim 0 We have lim x x x 1 x 1 Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial. 1 a =0 and a b 0 a = 1 and b = –1. Hence, a = 1 and b = – 1. (a) 2 Solution: (a) x x2 1 ax b 0 is The values of constants a and b so that lim x x 1 (a) a 0, b 0 (b) a 1, b 1 (c) a 1, b 1 (a) 1 Solution: (a) x x x x x x = lim E3 Solution: (b) x x x x ID Example: 14 lim x x x x = lim x x (d) e 4 (c) log 2 U Solution: (b) 1 2 (b) 60 (a) 0 (d) 0 1 x 2 1 2 2 x 0 equals [IIT 1978; Kurukshetra CEE 1998] (b) 2 3 3 (c) 0 (d) None of these Functions, Limits, Continuity and Differentiability Solution: (b) a 2x 3x a 2x 3x lim lim x a 3 a x 2 x x a 3 a x 2 x a 2x 3x a 2x 3x 75 3a x 2 x 3a x 2 x 3a x 2 x 2. lim x a 3( a 2 x 3 x ) 3 3 99 100 (a) Solution: (b) Example: 22 1 100 n lim n100 n (c) r 99 n r 1 n 100 199 299 3 99 .... n99 lim [EAMCET 1994] (b) 1 99 (d) n 99 lim 1 r n n n r 1 1 x 1 101 1 99 x 100 1. dx 100 100 0 0 x2 1 ax b 2 is The values of constants ‘a’ and ‘b’ so that lim x x 1 (b) a 1, b 1 (c) a 1, b 3 (d) a 2, b 1 ID (a) a 0, b 0 Solution: (c) = n100 n 60 199 299 3 99 .... n99 lim E3 Example: 21 x2 1 lim ax b 2 lim x 1 ax b 2 lim x (1 a) (1 b) 2. x x x x 1 n2 lim 3 n n D YG Example: 23 (a) Solution: (c) U Comparing the coefficient of both sides, 1 a 0 and 1 b 2 a 1, b 3 1 6 (c) 1 3 (d) lim : Students should remember that, n 1 2 2 n and lim 1 2 1 n2 n n 2 3 ...... ST n 1 n 2 lim n U n 1. 3 n is equal to 1 n2 (b) (a) 0 [IIT 1984; DCE 2000] 1 2 (c) 1 2 (d) None of these Solution: (b) 2 n 1 n 1 n2 n 1 lim ...... lim lim . 2 2 2 2 n 1 n 2 n 1 n 2 2 1n 1 n n 1 n Example: 25 If f (x ) 2 x 3 2(2 x 1) , g(x ) and h(x ) 2 then lim [ f (x ) g(x ) h(x )] is x 3 x 3 x 4 x x 12 (a) – 2 Solution: (c) 1 3 1 1 1 2 n(n 1) (2n 1) n n 1 lim nlim 3 n 6 3 6n Note Example: 24 1 6 (b) [Rajasthan PET 1999, 2002] We have f (x ) g(x ) h(x ) x 2 4 x 17 4 x 2 lim [ f (x ) g(x ) h(x )] lim x 3 (c) (b) – 1 x x 12 2 ( x 3)( x 5 ) 2 . 3 )( x 4 ) 7 x 3 (x 2 7 x 2 8 x 15 x x 12 2 (d) 0 ( x 3)( x 5 ) ( x 3 )( x 4 ) 76 Functions, Limits, Continuity and Differentiability n! If lim n n n 1/n equal [Kurukshetra CEE 1998] (a) e 1/n Solution: (b) 1 e (b) n! Let P lim n n n 4 (c) (d) 1/n n 1 2 3 4 P lim ............. n n n n n n n 1 1 2 n 1 r lim log log ......... log log P lim log n n n n n n n n r 1 log P log x dx [x log x x] 1 0 [Karnataka CET 2000] (b) a 1 and b 1 (c) a 1 and b 2 x 3 (1 a) bx 2 ax (1 b) x3 1 2 lim [ x 3 (1 a) bx 2 ax (1 b)] 2 (x 2 1). lim 2 (ax b) 2 lim 2 x x x x 1 x 1 Comparing the coefficients of both sides, 1 a 0 and b 2 or a 1, b 2. lim (x 1)10 (x 2)10 ..... (x 100 )10 x 10 10 10 x (a) 0 is equal to (b) 1 (c) 10 1 x 10 1 x lim 10 Example: 29 2002] (x 1)10 (x 2)10 ...... (x 100 )10 D YG Solution: (d) lim x 10 10 10 x x Let f (x ) 4 and f ' (x ) 4 , then lim x 2 (a) 2 Solution: (c) (d) a 1 and b 2 ID Example: 28 1. e x3 1 (ax b) 2, then If lim 2 x x 1 (a) a 1 and b 1 Solution: (c) (1) P U Example: 27 1 0 E3 log P 4 60 Example: 26 (b) – 2 2 f (x ) 2 f (2) xf (2) 2 f (2) (x 2) U y lim x 2 y 2 lim ST x 2 f (x ) f (2) f (2) x 2 (d) 100 10 100 ... 1 x 100. 10 10 x 10 1 10 x 10 2 1 x xf (2) 2 f (x ) equals x 2 xf (2) 2 f (x ) y lim x 2 x 2 [Rajasthan (c) – 4 xf (2) 2 f (2) 2 f (2) 2 f (x ) y lim x 2 x 2 y lim 2 x 2 [AMU 2000] 2000; AIEEE (d) 3 f (2).( x 2) [ f (x ) f (2)] lim x 2 (x 2) x 2 y 2 lim f (x ) f (2) 8 4 4. x 2 (2) Trigonometric limits : To evaluate trigonometric limits the following results are very important. x x sin x tan x (i) lim (ii) lim 1 lim 1 lim x 0 x x x 0 0 0 x x sin x tan x sin 1 x x 1 lim x 0 x 0 x sin 1 x (iii) lim sin x 0 x 0 x 180 sin( x a) (vii) lim 1 x a x a (v) lim (ix) lim sin 1 x sin 1 a, | a | 1 x a (iv) lim x 0 tan 1 x x 1 lim x 0 x tan 1 x (vi) lim cos x 1 x 0 (viii) lim x a tan( x a) 1 x a (x) lim cos 1 x cos 1 a; | a | 1 x a Functions, Limits, Continuity and Differentiability 77 sin x cos x (xii) lim lim 0 x x x x sin 1 / x lim 1 x 1 / x (xi) lim tan 1 x tan 1 a; a x a (xiii) x lim (1 x ) tan 2 2 (a) Solution: (c) [IIT 1978, 84; Rajasthan PET 1997, 2001; UPSEAT 2003] x 1 (b) (c) 2 x lim (1 x ) tan , Put 1 x y as x 1, y 0 2 x 1 1 cos 2(x 1) x 1 lim x 1 (a) Exists and it equal [IIT 1998; UPSEAT 2001] ID Example: 31 (d) 0 E3 y (1 y ) 2 2 2 2 1 . Thus lim y tan lim. y 0 y 0 2 y tan 2 60 Example: 30 2 f (1) lim f (1 h) lim h 0 h 0 1 cos 2h sinh lim 2 h 0 h h D YG Solution: (d) U (b) Exists and it equals 2 (c) Does not exist because x 1 0 (d) Does not exist because left hand limit is not equal to right hand limit 2 1 cos( 2 h) sinh lim 2 2. h 0 h h limit does not exist because left hand limit is not equal to right hand limit. f (1) lim f (1 h) lim h 0 Example: 32 lim x 2 sin 3 x 10 3 (b) 2 sin 2 x sin 5 x 3 x 5 x U Solution: (a) (1 cos 2 x ) sin 5 x x 0 (a) h 0 x 0 lim x 0 x 2 x3 sin x 2 sin 3 x 3 x 5 x (a) 0 Solution: (a) Example: 34 lim lim x2 sin x 2 lim sin 3 x sin x = x x 0 lim x 0 1 3 x 0 (b) x 0 (a) Solution: (c) x3 = lim 2 sin 2 x x2. (c) 6 5 (d) 5 6 (d) 1 2 (d) 1 4 5 10 3x sin 5 x 5 x = 2. ... 3 3 sin 3 x 5x 3x ST Example: 33 lim 3 10 [MP PET 2000; UPSEAT 2000; Karmataka CET 2002] x 0 sin x 2 1 3 (c) 3 x 2 lim x = 1.0 = 0.. x = lim x 0 sin x 2 x 0 (b) 3 (c) 4 sin 3 x sin x sin 3 x sin x sin 3 x sin x lim.3 lim = lim = lim = 1.3 + 1 = 4. x 0 x 0 x x 0 3 x x 0 x x x 78 Functions, Limits, Continuity and Differentiability Solution: (b) Example: 36 Solution: (d) 1 x sin , x 0 If f (x ) , then lim f ( x ) = [IIT 1988; UPSEAT 1988; SCRA 1996] x x 0 0, x 0 (a) 1 (b) 0 (c) –1 (d) None of these 1 1 lim x sin lim x lim sin = 0 × (A number oscillating between – 1 and 1) = 0. x 0 x x x 0 x 0 sin[ x ] , [x ] 0 If f (x ) [ x ] , then lim f ( x ) equals x 0 0 , [x ] 0 60 Example: 35 [IIT 1985; Rajasthan PET 1995] (a) 1 (b) 0 (c) –1 (d) Does not exist In closed interval of x 0 at right hand side [x] =0 and at left hand side [ x ] 1. Also =0. E3 sin[ x ] , (1 x 0) Therefore function is defined as f (x ) [ x ] 0 , (0 x 1) sin[ x ] sin(1) sin 1c [x ] 1 Right hand limit = 0, Hence, limit doesn’t exist. tan x sin x lim x 0 x3 1 1 2 (a) (b) (c) 2 2 3 Left hand limit lim f (x ) lim [IIT 1974; Rajasthan PET 2000] (d) None of these U Example: 37 x 0 ID x 0 Example: 38 If f (x ) Solution: (d) D YG Solution: (a) x x sin x 2 sin 2 sin 2 2 1 1 tan x sin x sin x sin x cos x 2 sin x 2 lim ... lim lim lim x 0 x 0 x cos x x 2 4 2 x 0 x 0 x 3 cos x x3 x 3 cos x 2 sin(e x 2 1) , then lim f (x ) is given by x 2 log( x 1) (a) – 2 (b) –1 lim f (x ) lim x 2 x 2 x 2 sin(e 1) sin(e 1) lim. t 0 log( t 1) log( t 1) (c) 0 (d) 1 t (Putting x = 2 + t) U sin(e t 1) 1 t 1 sin(e t 1) e t 1 t lim ..... lim.. t t 0 x e t 1 e 1 1! 2! t log(1 t) 1 1 t 1 t 2 ..... 2 3 [ As t 0, e t 1 0 , ST = 1.1.1 = 1 Example: 39 lim x / 2 acot x acos x cot x cos x [Kerala (Engg.) 2001] (a) log a Solution: (a) sin(e t 1) 1] (e t 1) (b) log 2 acot x acos x lim x / 2 cot x cos x (c) a (d) log x cot x cos x 1 lim acos x a cot x cos x x / 2 a cot x cos x 1 1 log a log a. a cos( / 2) lim x / 2 cot x cos x Example: 40 If f (x ) sin x x3 2x cos x x2 1 tan x f (x ) x , then lim is x 0 x 2 1 [Karnataka CET 2002] Functions, Limits, Continuity and Differentiability (a) 3 Solution: (d) (b) –1 (c) 0 79 (d) 1 f (x ) x (x 1) sin x (x 3 2 x 2 ) cos x x 3 tan x x 2 sin x x 3 cos x x 3 tan x 2 x 2 cos x x sin x Example: 41 sin x lim sin x x cos x x tan x 2 cos x 0 0 0 2 1 1. x 0 x 3x x 3 f (x ) cot 1 1 3x 2 If 2 and g(x ) cos 1 1 x 1 x 2 3 (a) Solution: (d) x 2 (b) 2(1 a ) 2 3 x x 3 f (x ) cot 1 1 3 x 2 3 , then lim f (x ) f (a) , 0 a 1 is 2 x a g( x ) g(a) (c) 2(1 x ) 2 2 1 1 x and g(x ) cos 1 x 2 Put x tan in both equation ID 3 tan tan 3 1 f ( ) cot 1 cot tan 3 1 3 tan 2 3 2 [Orissa JEE 2003] 60 x 0 (d) 3 2 E3 f (x ) Hence, lim f ( ) cot 1 cot 3 3 f ( ) 3 2 2..….(i) 2 1 tan cos 1 (cos 2 ) 2 g ( ) 2 and g( ) cos 1 2 1 tan U ….. (ii) Example: 42 D YG f (x ) f (a) 1 1 1 3 f (x ) f (a) lim Now lim f (x ). 3 . g ( x ) g ( a ) x a x a x a g(x ) g(a) g (x ) 2 2 lim x a x a x 1 tan [1 sin x ] 2 lim is x 3 x 2 1 tan [ 2 x ] 2 (a) 1 8 (b) 0 [AIEEE 2003] (c) 1 32 (d) U x tan (1 sin x ) 4 2 lim ( 2 x ) 3 x Solution: (c) 2 ST 2 y y y y y tan (1 cos y ) tan.2 sin 2 tan sin 2 2. 2 1. 2 2 lim 1 Let x y, then y 0 lim = lim 3 3 y y 0 32 32 y 0 y 0 2 y (2 y ) (8 )y 2 2 Example: 43 If lim x 0 (a) 0 Solution: (d) lim n x 0 [(a n) nx tan x ] sin nx x2 0, where n is non-zero real number, then a is equal to (b) n 1 n (c) n (d) n 1 n sin nx tan x 1 . lim (a n)n 0 n [(a n)n 1] 0 (a n)n 1 a n . nx x 0 x n (3) Logarithmic limits : To evaluate the logarithmic limits we use following formulae 80 Functions, Limits, Continuity and Differentiability x2 x3 ............ to where 1 x 1 and expansion is true only if base is e. 2 3 (ii) lim log(1 x ) 1 x (iii) lim log e x 1 (iv) lim log(1 x ) 1 x (v) lim x 0 x 0 Example: 44 lim x e x 0 log a (1 x ) log a e, a 0, 1 x log e (1 2h) 2 log e (1 h) [IIT Screening 1997] h2 h 0 (a) –1 60 (i) log(1 x ) x (b) 1 (c) 2 (d) –2 (2h) (2h) (2h) ..... 2 h h h ...... 2 3 2 3 log e (1 2h) 2 log e (1 h) = lim lim 2 2 x a h0 h h 2 h 2 2h3 .... h0 Example: 45 lim x a h 2 = lim h 2 {1 2h ....} h2 h0 (a) –1 (b) 2 Let x – a = y, when x a, y 0, Example: 46 The given limit = lim y 0 lim log{1 y} 1. y log 10 (1 h) h D YG h 0 (c) 1 (a) 1 Solution: (b) lim h 0 Example: 47 (b) log10 e (c) log e 10 (d) None of these 1 log e (1 h). log10 e. h log e 10 If lim x 0 log( 3 x ) log( 3 x ) k , then the value of k is x (b) (a) 0 log( 3 x ) log( 3 x ) lim lim x 0 x 0 x U Solution: (c) (d) –2 U Solution: (c) = lim {1 2 h ....} 1. h 0 log{1 (x a)} (x a) 3 E3 = lim 2 ID Solution: (a) 3 lim 1 3 2 3 (c) (d) 2 3 1 ( x / 3) 3 x log log 3 x lim 1 ( x / 3) x 0 x x log 1 (x / 3) log 1 (x / 3) 1 1 2 lim . x 0 x x 3 3 3 ST x 0 [AIEEE 2003] (4) Exponential limits : (i) Based on series expansion : We use e x 1 x x2 x3 ............. 2! 3! To evaluate the exponential limits we use the following results – (a) lim x 0 ex 1 1 x (b) lim x 0 ax 1 log e a x (c) lim x 0 e x 1 x ( 0) (ii) Based on the form 1 : To evaluate the exponential form 1 we use the following results. (a) If lim f (x ) lim g(x ) 0 , then lim {1 f (x )} x a x a x a 1 / g( x ) e lim f (x ) x a g ( x ) , or Functions, Limits, Continuity and Differentiability when lim f (x ) 1 and lim g(x ) . Then lim { f (x )} lim x 0 1 (c) 2 2 lim x 0 (d) e x 1 e x 1 e x e x (e x 1) (e x 1) lim lim = lim = . x 0 x 0 x 0 x x x x e x (1 x ) x2 is [Karnataka CET 1995] (b) e x (1 x ) x2 x 0 x 0 e x 60 (b) x 0 lim x [MP PET 1994] The value of lim lim E3 x 0 (1 x = lim ax 1 (c) 1 x2 is equal to 1 x 1 1 2 x2 .....) (1 x ) 2! x 0 (a) 2 log e a Solution: (a) x x 0 (b) ax 1 1 x 1 ax 1 lim 1 x 1 x 0 (d) 1 4 1 x x2 x 2 ..... 2! 3! 4 ! = 1 1. = lim 2 x 0 2! 2 x D YG Example: 50 (e) lim 1 e x e x x (a) 0 Solution: (b) (d) lim (1 x )1 / x e ID lim = e x a x a U Example: 49 lim [1 f (x ) 1] 81 lim ( f ( x )1)g ( x ) g( x ) , if a 1 lim a x i.e., a , if a 1 and a 0 if a 1. x 0 , if a 1 (a) Solution: (d) 1 e x x x 0 Example: 48 x (c) lim 1 (b) lim (1 x )1 / x e Note : x a x a x a g( x ) . 1 log e a 2 1 x 1 1 x 1 (c) a log e 2 (d) None of these ax 1 (a x 1) 1 x 1 . 1 x 1 = lim x 0 x 0 1 x 1 x = lim a x 1 . lim 1 x 1 = (log e a). (2) = 2 log e a. = lim x 0 x x 0 x 3 The value of lim x x 1 U Example: 51 ST (a) e 4 Solution: (d) x 3 lim x x 1 x 2 x 2 is (b) 0 2 lim 1 x x 1 Example: 52 x 2 (b) e c / a 2 1 2. x x 1 1 1 2 x 2 lim 1 x x 1 2 = lim 1 x x 1 1 If a, b, c, d are positive, then lim 1 x a bx (a) e d / b (d) e 2 (c) 1 x 1 2.( x 2). 2 ( x 1) x 3 Alternative method : lim x x 1 [UPSEAT 2003] x 2 lim 2 e x x 1 ( x 2) e e 2 1 2 lim 1 1 x x x 2 1 x lim 2 1 x 1 x e 2. e2 c dx [EAMCET 1992] (c) e (c d ) /(a b) (d) e 82 Functions, Limits, Continuity and Differentiability 1 lim 1 x a bx c dx a bx 1 lim 1 x a bx Alternative method : e lim x x 0 Solution: (a) ed / b. (b) 1 Let y x x y 0 x2 x 0 11e 24 (1 x ) (d) None of these 1 log(1 x ) ex x 0 x 0 1 [DCE 2001] 11e 24 e 24 (c) 1 x2 x3 x .... x 2 3 e x 1 ex 2 is (b) 1/ x (c) e log y x log x ; lim log y lim x log x 0 log 1 lim x The value of lim (a) [Roorkee 1987] (1 x )1 / x e Example: 54 c dx 1 (a) 0 Solution: (b) 1 a bx c dx d 1 lim 1 e and lim x a bx b x a bx = ed / b 60 Example: 53 x lim x a bx c dx a bx E3 Solution: (a) x x2 1 ....... 2 3 e (d) None of these x x2 ...... 2 3 ee x2 x 0 (1 x )1 / x e equals x 0 x lim D YG Example: 55 ex 2 11 e 24 (a) / 2 Solution: (d) (1 e 1 (b) 0 1 [log(1 x )] ex e 1 x x x x x .... 2 3 4 2 2 3 x x x ..... 2 3 4 1! 3 4 [UPSEAT 2001] (c) 2 / e 1 x x x .... 2 3 4 e 2 (d) – e / 2 x x x .... 2 3 4 e.e 2 3 3 x x2 x3 ... ex 11 e 2 2 3 4 x .......... ... e 2 24 2! 2 U = 1 x) x .. U lim (1 x )1 / x e ID 2 1 x x2 x x2 x 11 2 x .......... e 1 ..... ..... ..... e 1 2! 2 2 2 24 3 3 ST ex 11 e 2 e 2 24 x.......... e (1 x )1 / x e e e 11 e x ... . lim lim lim x 0 x 0 2 x 0 24 2 x x Example: 56 m x lim cos = m m [AMU 2001] (a) 0 Solution: (d) (b) e m x x lim cos lim 1 cos 1 m m m m x lim 1 2 sin 2 m 2m m (c) 1/e m x lim 1 cos 1 m m x lim 2 sin2 m m 2m e e x sin 2m lim 2 x m 2m (d) 1 m 2 x2 4m 2 m e 2 lim x2 m 4 m e0 1. n(n 1) Example: 57 n2 n 1 lim 2 n n n 1 [AMU 2002] Functions, Limits, Continuity and Differentiability 83 (c) e 1 (b) e 2 (a) e (d) 1 n(n 1) Solution: (b) n(n 1) n(n 1) 1 lim n n(n 1) 1 n(n 1) 2 Alternative Method: lim 1 2 n n n 1 1 1 n ( n 1) e lim 1 e 2. n(n 1) n e 1 1 n ( n 1 ) n(n 1) lim 2 n(n 1) = e n n 2 n 1 e2. 60 n n 1 lim 2 n n n 1 2 (5) L’ Hospital’s rule : If f (x ) and g(x ) be two functions of x such that (i) lim f (x ) lim g(x ) 0 x a E3 x a (ii) Both are continuous at x a (iii) Both are differentiable at x a. (iv) f ' (x ) and g' (x ) are continuous at the point x a , then lim Note : The above rule is also applicable if x a lim f (x ) and lim g(x ) . x a x a f ' (x ) assumes the indeterminate form 0 or and f ' (x ), g' (x ) satisfy all the condition 0 g' (x ) U If lim ID x a f (x ) f ' (x ) provided that g' (a) 0 lim g(x ) x a g' (x ) f ' (x ) f ' (x ) to get, lim x a g' (x ) g' (x ) D YG embodied in L’ Hospital rule, we can repeat the application of this rule on = lim x a f "( x ). Sometimes it may be necessary to repeat this process a number of times till our goal g "( x ) of evaluating limit is achieved. Example: 58 lim x 0 1 cos mx 1 cos nx (b) n / m (c) U (a) m / n [Kerala (Engg.) 2002] ST mx 2 mx 2 sin sin 1 cos mx 2 2 lim lim lim mx x 0 x 0 1 cos nx x 0 2 nx 2 sin 2 2 Solution: (c) m2 (d) n2 2 m 2 x 2 1. 4 nx sin 2 nx 2 2 n2 m2 4 m2 m2. 2 2 2 1 2 n x n n Trick : Apply L-Hospital rule , 1 cos mx m sin mx m 2 cos mx m2 lim lim 2 2. x 0 1 cos nx x 0 n sin nx x 0 n cos nx n lim Example: 59 The integer n for which lim (cos x 1) (cos x e x ) xn x 0 (a) 1 (b) 2 is a finite non-zero number is (c) 3 [IIT Screening 2002] (d) 4 84 Functions, Limits, Continuity and Differentiability Solution: (c) n cannot be negative integer for then the limit 0 x e x cos x 1 e x cos x 2 Limit = lim 2 ( n 1 for then the limit = 0) lim 2 n 2 x 0 2 (x / 2) 2 x 0 x n 2 x 2 sin 2 1 e x sin x 1 which is finite. If n 4 , the limit is infinite. lim. So, if n 3 , the limit is 2 x 0 (n 2)x n 3 2(n 2) 60 1 1 Example: 61 lim log f (1 x )log f (1) lim f (1 x ) x e x 0 x e x 0 lim x 0 f (1) lim / 4 (a) Solution: (a) 1 (b) 1 / 2 2 2 2 x 3 a3 x 2 a2 2. 2 (a) 0 Solution: (d) Example: 63 lim x a x h h lim h 0 3x 00 form = xlim a 3a 2 3a2 3a . 2a 2 [Roorkee 1983] (b) 1 / 2 h U x h x = lim h h 0 lim h 0 (By ‘L’ Hospital rule) = (d) (a) 1 / 2 x Solution: (a) (d) None of these (c) 2a 2 2x x e6 / 3 e2. (By ‘L’ Hospital rule) (b) Not defined x 3 a3 x 2 a2 f (1) f (1) (c) 1 D YG x a 1 e [IIT Screening 1997; AMU 1997] cos sin sin cos 0 ( 0 form) = lim / 4 / 4 1 lim / 4 lim f (1 x ) / f (1 x ) 1 sin cos / 4 2 = Example: 62 1 U Solution: (c) (d) e 3 (c) e 2 E3 (b) e 1 / 2 (a) 1 [IIT Screening 2002] ID Example: 60 f (1 x ) x Let f : R R be such that f (1) 3 and f (1) 6. Then lim equals x 0 f (1) x h h x x h x x h x (c) Zero = lim h 0 (d) None of these (x h) x h( x h = lim x) h 0 h h( x h = x) 1 2 x ST Trick : Applying ‘L’ Hospital’s rule, [Differentiating Nr and Dr with respect to h] 1 2 x h We get, lim h 0 1 Example: 64 lim 1 2 x. [MP PET 2001] 2 2 lim sin 2 sin 2 (a) 0 Solution: (d) 0 (b) 1 sin 2 sin 2 2 2 (c) sin (d) sin 2 2 = lim sin( ) sin( ) = lim sin( ) lim sin( ) = lim sin( ) = sin 2 . 0 ( ) ( ) ( ) 2 ( ) ( ) Trick : By L’ Hospital’s rule, lim 2 sin cos sin 2 . 2 2 Functions, Limits, Continuity and Differentiability [IIT 1971] (b) 1/3 Solution: (c) 2 tan 2 x 1 tan 2 x x 1 lim lim 2 x . sin x x 0 3 x sin x x 0 3 2 x Example: 66 If G(x ) 25 x 2 , then lim x 1 (a) 1/24 Solution: (d) (c) 1/2 G( x ) G(1) equals x 1 (c) 24 G(x ) G(1) 25 x 2 24 lim x 1 x 1 x 1 x 1 x 1 x 1 24 25 x 2 (d) None of these [Multiply both numerator and denominator by ( 24 25 x 2 )] 1 24 G(x ) 1(2 x ) lim x 1 1 2 25 x 2 1 ID Alternative method: By L'-Hospital rule, lim x 1 Example: 67 [IIT 1983] (b) 1/5 lim lim (d) 0 60 (a) 2/3 E3 Example: 65 85 tan 2 x x equals lim x 0 3 x sin x If f (a) 2, f ' (a) 1, g(a) 1, g' (a) 2, then lim x a 24 g(x ) f (a) g(a) f (x ) equals x a Solution: (c) Example: 68 (b) (d) (c) 3 Applying L – Hospital's rule, we get, lim x a 1 3 g(x ) f (a) g(a) f (x ) g(x ) f (a) g(a) f (x ) lim x a x a 1 g(a) f (a) g(a) f (a) 2 2 1 (1) 3. (1 x )n 1 x 0 x [Kurukshetra CEE 2002] lim (a) n (b) 1 (c) –1 (d) None of these (1 nx n C 2 x 2 ...... higher pow ers of x to x n ) 1 n x 0 x lim U Solution: (a) 1 3 D YG (a) –3 U [IIT 1983; Rajasthan PET 1990; MP PET 1995; DCE 1999; Karnataka CET 1999, 2003] Trick : Apply L- Hospital rule. lim sin x log(1 x ) x 0 x2 ST Example: 69 (a) 0 Solution: (c) is equal to [Roorkee 1995] (b) 1 2 (c) 1 2 (d) None of these 1 1 sin x 1 (1 x )2 1 x lim x 0 2x 2 2 cos x Apply L- Hospital rule, we get, lim x 0 3 5 2 3 4 x x x ...... x x x x ...... 3! 5! 2 3 4 sin x log(1 x ) lim lim Alternative method : lim 2 2 2 x 0 x 0 x 0 x x x 3 5 2 3 sin x x x x ...... and log (1 x ) x x x........ 3 ! 5 ! 2 3 86 Functions, Limits, Continuity and Differentiability x2 1 1 x4 x3 .... 1 2 3! 3 4 Hence, lim . x 0 2 x2 xe x log(1 x ) 2 3 (a) Solution: (d) equals x2 x 0 [Rajasthan PET 1996] 1 3 (b) Let y lim (c) xe x log(1 x ) x 0 x 1 2 0 form 0 2 Applying L–Hospital's rule, y lim e x xe x x 0 1 1 x 0 form 0 (d) 3 2 60 lim E3 Example: 70 2x 1 x 1 1 3 x x e e xe lim [1 1 0 1] 2 x 0 2 2 (1 x ) x 0 2 sin 1 x tan 1 x lim x3 x 0 (a) 0 Solution: (d) (b) 1 sin 1 x tan 1 x lim is equal to 3 x 0 (c) – 1 0 form 0 [Rajasthan PET 2000] (d) 1 2 U Example: 71 ID y lim Example: 72 D YG x Applying L-Hospital’s rule, 1 1 2 1 x2 0 1x lim form x 0 0 3x 2 1 2 x 2x 1 2 (1 x 2 ) 3 / 2 (1 x 2 ) 2 1 1 2 lim lim . 2 3 / 2 2 2 x 0 x 0 6x 6 (1 x ) (1 x ) 2 1 log x x lim x 1 1 2x x 2 = U (a) 1 (b) – 1 Applying L-Hospital’s rule, lim ST Solution: (d) x 1 1 log x x 1 2x x 2 Again applying L-Hospital’s rule, we get lim Example: 73 lim 4 x 9x x 0 x (4 x 9 x ) x 1 y lim x 0 (c) 0 (d) 1 2 1 1 1x lim x lim x 1 2 2 x x 1 2 x ( x 1) 1 1 4x 2 2 2 (a) log 3 Solution: (a) [Karnataka CET 2000] 4x 9x x (4 x 9 x ) [EAMCET 2002] 1 3 log 2 2 (b) (c) 1 3 log 2 2 3 (d) log 2 0 form 0 Using L-Hospital’s rule, y lim x 0 2 log 4 log 4 9 log 9 log 4 log 9 3 y y x x x 2 2 9 ) x (4 log 4 9 log 9) x (4 x x 2 log 2. 3 Functions, Limits, Continuity and Differentiability (a) 1 f (a) g(x ) f (x ) g(a) x a 0 form 0 Using L-Hospital’s rule, lim x a Example: 75 The value of lim 2 x 3 x 7 x 2 49 f (a) g(x ) f (x ) g(a) f (a) g (a) f (a) g(a) 2 (1) 1 (3) 1. 10 is 2 9 (a) [MP PET 2003] (b) 0 Solution: (d) Applying L-Hospital’s rule, lim Example: 76 Let x 7 f (a) g(a) k lim x a 1 56 (c) (d) 1 56 1 1 1 1 2 x 3 lim. x 7 4 x x 3 2x 4.7 7 3 56 derivatives f n (a) , g n (a) exist and are not equal for some n. If f (a) g(x ) f (a) g(a) f (x ) g(a) 4 , then the value of k is g(x ) f (x ) (b) 2 k g(x ) k f (x ) 4 g(x ) f (x ) (c) 1 [AIEEE 2003] (d) 0 U lim x a 2 49 n th and their (a) 4 Solution: (a) (d) – 1 60 lim x a (c) – 5 E3 Solution: (a) (b) 6 [Karnataka CET 2003] ID Example: 74 87 f (a) g(x ) f (x ) g(a) If f (a) 2 , f (a) 1 , g(a) 3 , g (a) 1 , then lim x a x a Example: 77 D YG g (x ) f (x ) By L-Hospital’ rule, lim k 4 , k 4. x a g (x ) f (x ) The value of lim x 0 sec 2 t dt is x sin x x2 0 (a) 3 (b) 2 d dx lim x2 sec 2 t dt 0 U Solution: (c) x 0 d (x sin x ) dx lim x 0 (c) 1 sec 2 x 2. 2 x sin x x cos x (d) 0 (By L' –Hospital's rule) 2 sec 2 x 2 2 1 1. 1 1 sin x cos x x ST = lim [AIEEE 2003] x 0 Example: 78 3 sin x 3 cos x lim x / 6 6x (a) Solution: (b) [EAMCET 2003] (b) 3 1 (c) 3 3 Using L–Hospital’s rule, lim x / 6 3 cos x 3 sin x 6 3. 3 1 3. 2 2 1. 6 3 (d) 1 3 88 Functions, Limits, Continuity and Differentiability Given that f ' (2) 6 and f (1) 4 , then lim h 0 (a) Does not exist lim h 0 f (2 h 2 h 2 ) f (2) f (h h 1) f (1) 2 lim h 0 f (h h 2 1) f (1) 3 2 (c) f (2h 2 h 2 )(2 2h) f (h h 2 1)(1 2h) 3 2 62 3. 4 1 [IIT Screening 2003] (d) 3 ST U D YG U ID E3 Solution: (d) (b) f (2 h 2 h 2 ) f (2) 60 Example: 79
