Chapter 2.2 Functions, Limits, Continuity and Differentiability PDF

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 53  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 10 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  a00  a. lim  lim e f x  dx 2  ex  f x  d 00 d d  2 x x ax 2  bx  c Example: 13 1  x 2  4  x  x  2  4  x 1 1 4x  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 n2 (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 x2  (c) 1   lim   1  x    (d) 0 444 3. 22 [Rajasthan PET 1988]   x 1 x  1x   lim   x 0  1  x  1  x  1  x  1  x   x 1 x  1x  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 1n 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  1x  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 1x  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. 10 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 62 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