Chapter 2.3 Continuity PDF

60 Functions, Limits, Continuity and Differentiability 97 Introduction. E3 The word ‘Continuous’ means without any break or gap. If the graph of a function has no break, or gap or jump, then it is said to be continuous. A function which is not continuous is called a discontinuous function. continuou...

60 Functions, Limits, Continuity and Differentiability 97 Introduction. E3 The word ‘Continuous’ means without any break or gap. If the graph of a function has no break, or gap or jump, then it is said to be continuous. A function which is not continuous is called a discontinuous function. continuous. Similarly tan x , cot x , sec x , 1 etc. are also discontinuous function. x U Continuous function Y ID While studying graphs of functions, we see that graphs of functions sin x , x, cos x , ex etc. are continuous but greatest integer function [x] has break at every integral point, so it is not Y (0, 1) – – /2  D YG –2 2 X O /2 (0,– 1) Y Discontinuous function Y 3 2 1 f (x) = x X X X 0 X f(x)= 1/x y = sin x Y X’ – 3 O 1 – –1 –1 2 – – 2 3 Y’ X 2 3 y = [x] U 2.3.1 Continuity of a Function at a Point. A function f (x ) is said to be continuous at a point x  a of its domain iff lim f (x )  f (a). i.e. a x a ST function f (x ) is continuous at x  a if and only if it satisfies the following three conditions : (1) f (a) exists. (‘a’ lies in the domain of f) f (x ) exist i.e. lim f (x )  lim f (x ) or R.H.L. = L.H.L. (2) lim x a x a x a (3) lim f (x )  f (a) (limit equals the value of function). x a Cauchy’s definition of continuity : A function f is said to be continuous at a point a of its domain D if for every   0 there exists   0 (dependent on  ) such that | x  a |   | f (x )  f (a)|  . Comparing this definition with the definition of limit we find that f (x ) is continuous at x  a if lim f (x ) exists and is equal to f (a) i.e., if lim f (x )  f (a)  lim f (x ). x a x a x a  98 Functions, Differentiability Limits, Continuity and Heine’s definition of continuity : A function f is said to be continuous at a point a of its domain D, converging to a, the sequence  a n  of the points in D converging to a, the sequence < f (a n )  converges to f (a) i.e. lim a n  a  lim f (a n )  f (a). This definition is mainly used to prove the discontinuity to a function. 60 Note :  Continuity of a function at a point, we find its limit and value at that point, if these two exist and are equal, then function is continuous at that point. Formal definition of continuity : The function f (x ) is said to be continuous at x  a, in its domain if for any arbitrary chosen positive number   0 , we can find a corresponding number 2.3.2 Continuity from Left and Right. Function f (x ) is said to be (1) Left continuous at x = a if lim f (x )  f (a) x a  0 x a  0 ID (2) Right continuous at x  a if Lim f (x )  f (a). E3  depending on  such that| f (x )  f (a)|   x for which 0 | x  a |  . Thus a function f (x ) is continuous at a point x  a if it is left continuous as well as right Solution: (d)  x  , x  3  x  3 is continuous at x = 3, then  = If f(x) =  4 , 3 x  5 , x  3  D YG Example: 1 U continuous at x  a. (a) 4 (b) 3 L.H.L. at x = 3, lim f (x )  lim (x   ) (c) 2 = lim (3  h   ) = 3   R.H.L. at x = 3, lim  f (x )  lim  (3 x  5) = lim {3(3  h)  5} = 4 x 3 x 3 x 3 x 3 (d) 1 …..(i) h 0 …..(ii) h 0 Value of function f (3)  4 …..(iii) For continuity at x = 3 Limit of function = value of function 3 +  = 4   = 1. 1   x sin , x  0 If f (x )   is continuous at x = 0, then the value of k is x  k , x 0 U Example: 2 (a) 1 (b) –1 (c) 0 (d) 2 If function is continuous at x = 0, then by the definition of continuity f (0)  lim f (x ) ST Solution: (c) [MP PET 1999; AMU 1999; Rajasthan PET 20 x 0 1  since f (0 )  k. Hence, f (0 )  k  lim (x )  sin  x 0 x   k = 0 (a finite quantity lies between –1 to 1)  k = 0. Example: 3 Solution: (c)  2 x  1 when x  1  when x  1 is continuous at x =1, then the value of k is If f (x )   k 5 x  2 when x  1  (a) 1 (b) 2 Since f (x ) is continuous at x = 1,  (c) 3 lim f (x )  lim f (x )  f (1) x 1  Now h 0 (d) 4 …..(i) x 1  lim f (x )  lim f (1  h) = lim 2(1  h)  1  3 i.e., lim f ( x )  3  x 1  [Rajasthan PET 2001] h 0 x 1 Similarly, lim f (x )  lim f (1  h) = lim 5(1  h)  2 i.e., lim f (x )  3   h 0 x 1 h 0 Functions, Limits, Continuity and Differentiability 99 x 1 So according to equation (i), we have k = 3.  1 sin  , x  0 The value of k which makes f (x )    x  continuous at x = 0 is [Rajasthan PET 1993; UPSEAT 1995]  k , x 0  (a) 8 Solution: (d) (b) 1 We have lim f (x )  lim sin x 0 x 0 (c) –1 (d) None of these 60 Example: 4 1 = An oscillating number which oscillates between –1 and 1. x Hence, lim f ( x ) does not exist. Consequently f (x ) cannot be continuous at x  0 for any value of k. x 0  mx 2 , x  1 The value of m for which the function f (x )   is continuous at x  1 , is   2 x, x  1 (a) 0 (b) 1 (c) 2 LHL = lim f (x )  lim m(1  h)2  m x 1  h 0 (d) Does not exist ID Solution: (c) E3 Example: 5 RHL = lim f (x )  lim 2(1  h)  2 and f (1)  m x 1 h 0 Function is continuous at x  1 ,  LHL = RHL = f (1)  (cos x )1 / x , x  0 If the function f (x )   is continuous at x  0 , then the value of k is  ,x  0 k D YG Example: 6 U Therefore m  2. (a) 1 Solution: (a) (b) –1 lim (cos x )1 / x  k  lim x 0 x 0 (c) 0 (d) e 1 1 1 lim log cos x  log k  lim  0  log e k  k  1. log(cos x )  log k  lim x 0 x x 0 x x 0 x 2.3.3 Continuity of a Function in Open and Closed Interval. Open interval : A function f (x ) is said to be continuous in an open interval (a, b) iff it is U continuous at every point in that interval. Note :  This definition implies the non-breakable behavior of the function f (x ) in the ST interval (a, b). Closed interval : A function f (x ) is said to be continuous in a closed interval [a, b] iff, (1) f is continuous in (a, b) (2) f is continuous from the right at ‘a’ i.e. lim f (x )  f (a) x a (3) f is continuous from the left at ‘b’ i.e. lim f (x )  f (b). x b 100 Functions, Differentiability Example: 7 Limits, Continuity and  2 ,  x  a 2 sin x  If the function f (x )   x cot x  b ,   b sin 2 x  a cos 2 x ,   0x  4  2  4 x  , is continuous in the interval [0, ] then the values 2  x  of (a, b) are Solution: (b) (b) (0, 0) Since f is continuous at x    4  b  4  a 2 2 sin  4 1  b  a2 2.  b  a2 2  2 ;  f     2    f (x )  lim f   h  h 0  2  0 lim x  2    b sin 2  a cos 2  lim (  h) cot(  h)  b   b. 0  a(1)  0  b  a  b. 2 2 h 0  2 2  ID  (d) (1, –1)          ;  f    f   h   f   h   (1)  b    0   a 2 2 sin  0  4 4 4 4 4 4 4       h 0   h 0   Also as f is continuous at x   (c) (–1, 1) E3 (a) (–1, –1) 60 [Roorkee 1998] Hence (0, 0) satisfy the above relations. Solution: (c) U So, D YG Example: 8 x  1  sin 2 for    x  1  for 1  x  3 is continuous in the interval (, 6) then the values of a If the function f (x )   ax  b  x for 3  x  6 6 tan 12  and b are respectively [MP PET 1998] (a) 0, 2 (b) 1, 1 (c) 2, 0 (d) 2, 1  The turning points for f (x ) are x  1, 3.       lim f (x )  lim f (1  h) = lim 1  sin (1  h) = 1  sin  0  = 2 h 0  h 0 2 2     x 1  Similarly, lim f (x )  lim f (1  h) = lim a(1  h)  b = a + b h 0 x 1 h 0  f (x ) is continuous at x  1, so lim f (x )  lim f (x )  f (1)   x 1 x 1 U  2ab..…..(i) Again, lim  f (x )  lim f (3  h) = lim a(3  h)  b = 3a  b and lim f (x )  lim f (3  h) = lim 6 tan  x 3 h 0 h 0 x 3 h 0 h 0  12 (3  h) = 6 f (x ) is continuous in (, 6) , so it is continuous at x  3 also, so lim f (x )  lim f (x )  f (3)   ST x 3 x 3  3a  b  6..….(ii) Solving (i) and (ii) a = 2, b = 0. Trick : In above type of questions first find out the turning points. For example in above question they are x = 1,3. Now find out the values of the function at these points and if they are same then the function is continuous i.e., in above problem.   1  sin 2 x ;    x  1,  f (x )   ax  b ; 1 x 3  x ; 3x 6 6 tan 12  f (1)  2 f (1)  a  b, f (3)  3 a  b f (3)  6 Which gives 2  a  b and 6  3a  b after solving above linear equations we get a  2, b  0.   when 0  x   x sin x , 2 then If f (x )      sin(  x ), when  x   2 2 (a) f (x ) is discontinuous at x   (b) f (x ) is continuous at x  2 (c) f (x ) is continuous at x  0 lim  f (x )  Solution: (a) x   2 , lim f (x )   2 x   2 [IIT 1991]    and f   . ` 2 2 2 2 x   2 2 (a) 8 (b) –8 ID   1  cos 4 x , when x  0  x2  If f (x )   a , when x  0 is continuous at x = 0, then the value of ‘a’ will be  x  , when x  0  (16  x )  4  (c) 4 [IIT 1990; AMU 2 (d) None of these  2 sin 2 2 x  4  8 and lim f (x )  lim [( 16  x )  4 ]  8 lim f (x ) lim  2  x 0  x 0   (2 x ) x 0  x 0   Solution: (a) D YG Hence a  8. U Example: 10   E3 x  2 None of these (d) Since lim  lim f (x ) ,  Function is discontinuous at x   60 Example: 9 Functions, Limits, Continuity and Differentiability 101 2.3.4 Continuous Function. (1) A list of continuous functions : Function f(x) (i) Constant K Interval in which f(x) is continuous (–, ) (–, ) (iii) x–n (n is a positive integer) (–, ) – {0} U (ii) xn, (n is a positive integer) (iv) |x – a| p(x )  a0 x n  a1 x n 1  a2 x n  2 ........  an ST (v) (vi) p( x ) , where p(x) and q(x) are polynomial in q( x ) (–, ) (–, ) (–, ) – {x : q(x) = 0} x (vii) sin x (–, ) (viii) cos x (–, ) (ix) tan x (–, ) – {(2n + 1)/2 : n  I} (x) cot x (xi) sec x (–, ) – {n  : n  I} (–, ) – {(2n  1) /2 : n  I} 102 Functions, Differentiability (xii) cosec x Limits, Continuity and (–, ) – {n  : n  I} (–, ) (xiii) e x (0, ) (xiv) log e x 60 (2) Properties of continuous functions : Let f (x ) and g(x ) be two continuous functions at x  a. Then (i) cf (x ) is continuous at x = a, where c is any constant E3 (ii) f (x )  g(x ) is continuous at x  a. (iii) f (x ). g(x ) is continuous at x  a. (iv) f (x ) / g(x ) is continuous at x  a , provided g(a)  0. ID Important Tips  A function f (x ) is said to be continuous if it is continuous at each point of its domain.  A function f (x ) is said to be everywhere continuous if it is continuous on the entire real line R i.e. (, ). eg. U polynomial function e x , sin x, cos x, constant, x n , | x  a | etc. Integral function of a continuous function is a continuous function.  If g(x) is continuous at x = a and f(x) is continuous at x = g(a) then (fog) (x) is continuous at x  a.  If f(x) is continuous in a closed interval [a, b] then it is bounded on this interval.  If f(x) is a continuous function defined on [a, b] such that f(a) and f(b) are of opposite signs, then there is atleast one value of x for which f(x) vanishes. i.e. if f(a)>0, f(b) < 0  c  (a, b) such that f(c) = 0. D YG   If f(x) is continuous on [a, b] and maps [a, b] into [a, b] then for some x  [a, b] we have f(x) = x. (3) Continuity of composite function : If the function u  f (x ) is continuous at the point x  a, and the function y  g(u) is continuous at the point u  f (a) , then the composite function y  (gof )(x )  g( f (x )) is continuous at the point x = a. U 2.3.5 Discontinuous Function. ST (1) Discontinuous function : A function ‘f’ which is not continuous at a point x  a in its domain is said to be discontinuous there at. The point ‘a’ is called a point of discontinuity of the function. The discontinuity may arise due to any of the following situations. (i) lim f (x ) or lim f (x ) or both may not exist x a x a (ii) lim f (x ) as well as lim f (x ) may exist, but are unequal. x a x a (iii) lim f (x ) as well as lim f (x ) both may exist, but either of the two or both may not be equal x a x a to f (a). Important Tips  A function f is said to have removable discontinuity at x = a if lim f (x )  lim f (x ) but their common value is not equal x a to f(a). x a Functions, Limits, Continuity and Differentiability 103 Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.  If lim f (x ) does not exist, then we can not remove this discontinuity. So this become a non-removable discontinuity  or essential discontinuity. If f is continuous at x = c and g is discontinuous at x = c, then x a (a) f +g and f – g are discontinuous (b) f.g may be continuous If f and g are discontinuous at x = c, then f + g, f – g and fg may still be continuous.  Point functions (domain and range consists one value only) is not a continuous function. The points of discontinuity of y  u u  2 where u  1 (b) , 1,  2 2 1 (a) , 1, 2 2 1 is x 1 1 ,  1, 2 2 (c) when u   2  1 1 1  1  x  2.  2  x  , when u  1  x 1 2 x 1 The function f (x )  1 , 1, 2. 2 U Hence, the composite y  g( f ( x )) is discontinuous at three points  Example: 12 (d) None of these 1 is discontinuous at the point x  1. The function x 1 1 1  y  g(x )  2 is discontinuous at u  2 and u  1 u  u  2 (u  2) (u  1) The function u  f (x )  ID Solution: (a) 1 2 E3 Example: 11 60  log(1  ax )  log(1  bx ) is not defined at x  0. The value which should be assigned to x D YG f at x = 0 so that it is continuous at x = 0, is (b) a  b (a) a  b Solution: (b) (c) log a  log b (d) log a  log b Since limit of a function is a  b as x 0 , therefore to be continuous at x  0 , its value must be a  b at x  0  f (0)  a  b. Example: 13 x2  4x  3  If f (x )   x 2  1 , for x  1 , then 2 , for x  1  lim f ( x )  2 x 1  U (a) [IIT 1972] (b) (c) f (x ) is discontinuous at x  1 f (1)  2, f (1)  lim ST Solution: (c) f (1)  lim x 1 Example: 14 x2  4x  3 x 1 x2  4x  3 x2 1 x2 1  lim x 1 (d) None of these (x  3)  1 (x  1)  1  f (1)  f (1). Hence the function is discontinuous at x  1.  x  1, x  0  1 If f (x )   , x  0 , then  42  x , x  0 (a) lim f (x )  1 x 0  (c) f (x ) is discontinuous at x  0 Solution: (c) lim f ( x )  3 x 1  [Roorkee 1988] (b) lim f (x )  1 x 0  (d) None of these Clearly from curve drawn of the given function f (x ) , it is discontinuous at x  0. 104 Functions, Differentiability Limits, Continuity and y y = x2, x > 0 (0,1/4) x' x O 60 (0,–1) y = x-1, x < 0   x 0 6 , then the values of a and b if f is continuous at x  0 , are x 0 0x respectively (b) a  (1 | sin x |) | sin x|  f (x )   b tan 2 x   e tan 3 x  2 2/3 ,e 3 (c) 3 3/2 ,e 2 (d) None of these      x  0 6 x 0   0 x   6 ; ; ; D YG Solution: (b) 2 3 , 3 2 6 U (a)  ID Example: 15 a  (1 | sin x |) | sin x| ,   Let f (x )   b ,  tan 2 x  e tan 3 x ,   E3 y' For f (x ) to be continuous at x  0  a lim | sin x|  | sin x| a  lim f (x )  f (0 )  lim f (x )  lim (1 | sin x |) | sin x|  e x 0 x 0 x 0 x 0  tan 2 x   tan 3 x .2 x  .3 x   2x   3x  Now, lim e tan 2 x / tan 3 x  lim e  U x 0   ea  b  e2 / 3  a   ea  lim e 2 / 3  e 2 / 3. x 0  2 and b  e 2 / 3. 3 x Let f (x ) be defined for all x  0 and be continuous. Let f (x ) satisfy f    f (x )  f (y) for all x, y and y ST Example: 16 x 0      f (e )  1, then (a) f (x )  In x Solution: (a) [IIT 1995] (b) f (x ) is bounded 1 (c) f   0 as x 0 (d) xf (x ) 1 as x 0 x Let f (x )  In (x ), x  0 f (x )  In (x ) is a continuous function of x for every positive value of x. x x f    In    In (x )  In (y )  f (x )  f (y ). y y Functions, Limits, Continuity and Differentiability 105 Example: 17    Let f (x )  [x ] sin   , where [.] denotes the greatest integer function. The domain of f is ….. and  [x  1]  the points of discontinuity of f in the domain are (b) {x  R | x  [1, 0)}, I  {0} (c) {x  R | x  [1, 0)}, I  {0} (d) None of these 60 Solution: (c) (a) {x  R | x  [1, 0)}, I  {0} Note that [ x  1]  0 if 0  x  1  1 i.e. [ x  1]  0 if 1  x  0. E3 Thus domain of f is R  [1, 0)  {x  [1, 0)}    We have sin  is continuous at all points of R  [1, 0) and [x ] is continuous on R  I, where I  [ x  1]  denotes the set of integers. Thus the points where f can possibly be discontinuous are……, 3,  2,  1, 0 1, 2,........... But for ID     is defined. 0  x  1, [x ]  0 and sin  [ x  1]  Therefore f ( x )  0 for 0  x  1. U Also f (x ) is not defined on 1  x  0. Therefore, continuity of f at 0 means continuity of f from right at 0. Since f is continuous from Example: 18 f (0 ) is D YG right at 0, f is continuous at 0. Hence set of points of discontinuities of f is I  {0}. If the function f (x )  2 x  sin 1 x 2 x  tan 1 x (a) 2 Solution: (b) , (x  0) is continuous at each point of its domain, then the value of (b) 1/3  2 x  sin 1 x    f (0) f (x )  lim  x 0  2 x  tan 1 x    [Rajasthan PET 2000] (c) 2/3 (d) – 1/3 0  ,  form  0   ST U   1 2    2  1x  2 1 1    Applying L-Hospital’s rule, f (0 )  lim x 0  2 1 3 1   2   1 x2   sin 1 x 2 1 1 2 x  sin x x  lim  . Trick : f (0) = lim x 0 2 x  tan 1 x x 0 2 1 3 tan 1 x 2 x Example: 19 1 2   x   2 sin x , 2    f ( x )  A sin x  B ,   x  , is continuous everywhere The values of A and B such that the function  2 2    cos x , x  2 are [Pb. CET 2000] 106 Functions, Differentiability Limits, Continuity (a) A  0, B  1   lim (2 sin x )  For continuity at all x  R , we must have f     lim ( A sin x  B) x ( / 2)  2  x ( / 2)  2  A  B …..(i)   and f    lim  ( A sin x  B)  lim (cos x ) x ( / 2)  2  x ( / 2)  0  AB ….(ii) From (i) and (ii), A  1 and B  1. If f (x )  x 2  10 x  25 for x  5 and f is continuous at x  5, then f (5 )  x 2  7 x  10 (a) 0 [EAMCET 2001] E3 Example: 20 (d) A  1, B  0 (c) A  1, B  1 (b) A  1, B  1 60 Solution: (c) and (b) 5 (c) 10 (d) 25 (x  5 )2 5 5  0. x 5 ( x  2)(x  5 ) 52 x 2  10 x  25  lim Solution: (a) f (5 )  lim f ( x )  lim Example: 21 In order that the function f (x )  (x  1)cot x is continuous at x  0 , f (0 ) must be defined as x 5 x  7 x  10 ID x 5 2 1 (a) f (0 )  e x 0 1     lim (1  x ) x  x 0     x cot x  lim  x   1  x 0  tan x      lim (1  x ) x   e1  e. x 0     D YG x 0 The function f (x )  sin | x | is (a) Continuous for all x [DCE 2002] (b) Continuous only at certain points (c) Differentiable at all points Solution: (a) ST f (x ) is continuous at x  (b) 1  , then 2 If f (x )  (a) Solution: (d) 1 4 [Rajasthan PET 2002] (c) 0 lim f (x )  f (0) or   lim x  / 2 x  / 2 Applying L-Hospital’s rule,   lim Example: 24 None of these   1  sin x , x    2 x 2 be continuous at x   , then value of  is If f (x )    2  , x  2  (a) –1 Solution: (c) (d) It is obvious. U Example: 23 (d) None of these For continuity at 0, we must have f (0 )  lim f ( x )  lim ( x  1) cot x Example: 22 (c) f (0 )  e U Solution: (c) (b) f (0)  0 [UPSEAT 2000; Haryana CEE 2001] x  / 2 (d) 2 1  sin x   2x 0  ,  form  0    cos x cos x  0.    lim x  / 2 2 2 2 x4 ; (x  0 ), is continuous function at x  0 , then f (0 ) equals sin 2 x (b)  1 4 (c) If f (x ) is continuous at x  0, then, f (0 )  lim f (x )  lim x 0 x 0 1 8 2 x4 sin 2 x [MP PET 2002] (d)  0  ,  form  0   1 8  1   2 x4  Using L–Hospital’s rule, f (0)  lim x 0 2 cos 2 x Example: 25     Functions, Limits, Continuity and Differentiability 107  1. 8 if x is rational  x If function f (x )   , then f (x ) is continuous at...... number of points 1  x if x is irrational  (a)  Example: 26 (c) 0 At no point, function is continuous.  1  2 2 x x  e The function defined by f (x )     k     1 , x  2 , is continuous from right at the point x = 2, then , x 2 k is equal to (b) 1/4     and f (2)  k (c) –1/4 U 1  f (x )   x 2  e 2  x   1 [Orissa JEE 2002] (d) None of these ID (a) 0 Solution: (b) (d) None of these E3 Solution: (c) (b) 1 60 [UPSEAT 2002] If f (x ) is continuous from right at x  2 then lim f (x )  f (2)  k x 2 1 1   2 2 (2  h)    k  k  lim f (2  h)  k  lim (2  h)  e  h 0  h 0   D YG 1    lim  x 2  e 2  x   x 2       k  lim 4  h 2  4 h  e 1 / h h 0 Example: 27 1 1 2 ST (a)  Solution: (c) x  2 sin 2 lim f (x )  lim x x  x  2 cos 2  2 sin 2  4  1  kx  1  kx  If f (x )   x  2x 2  3x  2  (a) – 4 Solution: (c) (b) 2 cos 2  At x   , f ( )   tan Example: 28  k  [4  0  0  e  ] 1  k  1 2 x 0 [Orissa JEE 2003] (c) – 1 x x x cos  sin cos 2 2 2  lim x x x x  cos  sin cos 2 2 2 (d) 1 x 2  lim tan    x  x x  4 2 2  1. , for  1  x  0 is continuous at x  0, then k  [EAMCET 2003] , for 0  x  1 (b) – 3 L.H.L.  lim  1. 4 1  sin x  cos x is not defined at x  . The value of f ( ), so that f (x ) is continuous 1  sin x  cos x U at x   , is The function f (x )   1 1  kx  1  kx k x (c) – 2 (d) – 1 108 Functions, Differentiability Limits, Continuity and R.H.L.  lim  (2 x 2  3 x  2)  2 x 0 Since it is continuous, hence L.H.L = R.H.L  k  2. The function f (x ) | x |  | x| is x [Karnataka CET 2003] (a) Continuous at the origin (b) Discontinuous at the origin because |x| is discontinuous there | x| is discontinuous there x (d) Discontinuous at the origin because both |x| and | x | is continuous at x  0 and | x| is discontinuous at x  0. x ST U D YG U  f (x ) | x |  | x| is discontinuous at x  0 x ID Solution: (c) | x| are discontinuous there x E3 (c) Discontinuous at the origin because 60 Example: 29