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(2) KEY CONCEPTS (LIMIT) THINGS TO REMEMBER : 1. Limit of a function f(x) is said to exist as, x→a when Lim f (x) = Lim f (x) = finite quantity.. x →a −. 2.. x →a +. FUNDAMENTAL THEOREMS ON LIMITS: Let Lim f (x) = l & Lim g (x) = m. If l & m exists then : x →a. x →a. REMEMBER. (ii) Lim f(x). g(x) = l. m. (i). Lim f (x) ± g (x) = l ± m. (iii). f (x ) l Lim = , provided m ≠ 0 x →a g(g ) m. (iv). Lim k f(x) = k Lim f(x) ; where k is a constant.. (v). Lim f [g(x)] = f Lim g( x ) = f (m) ; provided f is continuous at g (x) = m. x →a x →a. 3.. x →a. x →a. x →a. Limit ⇒ x ≠ a x →a. x →a. For example Lim l n (f(x) = ln Lim f ( x ) l n l (l > 0). x →a x →a STANDARD LIMITS :. (a). sin x tan x tan −1 x sin −1 x = 1 = Lim = Lim = Lim x x x →0 x →0 x →0 x →0 x x [Where x is measured in radians]. (b). Lim (1 + x)1/x = e = Lim 1 + 1 x →0 x →∞ x. Lim. x. note however there Lim (1 – h)n = 0 h →0 n →∞. and Lim (1 + h )n → ∞ h →0 n →∞. (c). Lim f(x) = 1 and Lim φ (x) = ∞ , then ;. If. x →a. Lim [f ( x )]φ( x ) = e. x →a Lim φ ( x )[ f ( x ) −1] x →a. x →a. (d). Lim f(x) = A > 0 & Lim φ (x) = B (a finite quantity) then ;. If. x →a. x →a. Lim [f(x)] φ(x) = ez where z = Lim φ (x). ln[f(x)] = eBlnA = AB x →a. x →a. x Lim a − 1 = ln a (a > 0). In particular Lim e −1 = 1 x. (e). x →0. x →0. x. x. x −a = n a n −1 x − a x →a n. n. (f). Lim. 4.. SQUEEZE PLAY THEOREM : Limit Limit If f(x) ≤ g(x) ≤ h(x) ∀ x & Limit x → a f(x) = l = x → a h(x) then x → a g(x) = l.. 5.. INDETERMINANT FORMS : 0 ∞ , , 0 × ∞, 0°, ∞°, ∞ − ∞ and 1∞ 0 ∞ ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 2 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(3) Note : (i) We cannot plot ∞ on the paper. Infinity (∞) is a symbol & not a number. It does not obey the laws of elementry algebra. (ii) ∞+∞=∞ (iii) ∞ × ∞ = ∞ (iv) (a/∞) = 0 if a is finite a (v) is not defined , if a ≠ 0. 0 (vi) a b = 0 , if & only if a = 0 or b = 0 and a & b are finite. The following strategies should be born in mind for evaluating the limits: Factorisation Rationalisation or double rationalisation Use of trigonometric transformation ; appropriate substitution and using standard limits Expansion of function like Binomial expansion, exponential & logarithmic expansion, expansion of sinx , cosx , tanx should be remembered by heart & are given below :. 6. (a) (b) (c) (d). x ln a x 2 ln 2 a x 3ln 3a (i) a = 1 + + + + .........a > 0 1! 2! 3! x. x (ii) e = 1 +. x x2 x3 + + + ............ x ∈ R 1! 2! 3!. (iii) ln(1+ x) = x −. x 2 x3 x 4 + − + .........for − 1 < x ≤ 1 2 3 4. π π x3 x5 x7 + − + ... x ∈ − , (iv) sin x = x − 2 2 3! 5! 7! π π x2 x4 x6 + − + ...... x ∈ − , (v) cos x = 1 − 2 2 2! 4! 6!. (vi) tan x = x +. x 3 2x 5 + + ........ x ∈ 3 15. (vii) tan–1x = x −. π π − , 2 2. x3 x5 x7 + − + ....... 3 5 7. EXERCISE–I x2 − x Lim Q.1 x →1 x −1. Q.4. Q.7. 13. Q.2. Lim. x →1 5. x −7 x x −3 x. 100 k ∑ x −100 Lim K=1 x →1 x −1. 2 x + 3x1/ 3 + 5x1/ 5 Q.5 Lim 1/ 3 x →∞. Lim sec 4 x − sec 2 x x →0 sec 3x − sec x. q p − p, q ∈ N Q.8 Lim p x →1 1 − x 1 − x q . 3x − 2 + (2x − 3). Q.3. 2 Lim x − x.1nx + 1nx − 1 x →1 x −1. Q.6 Lim x→. 3π 4. 1 + 3 tan x 1 − 2 cos 2 x. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 3 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(4) Q.9. Find the sum of an infinite geometric series whose first term is the limit of the function f(x) = as x → 0 and whose common ratio is the limit of the function g(x) =. Q.10. tan x − sin x sin 3 x. 1− x as x → 1. (cos −1 x) 2. t −t Lim (x − l n cosh x) where cosh t = e + e .. x →∞. 2. cos −1 2 x 1 − x 2 [ x ]2 + 15[ x ] + 56 1 − sin 2x Lim Q.11 (a) Lim ; (b) Lim ; (c) π x→ 4 1 x →−7 sin( x + 7) sin( x + 8) 1 π − 4x x→ − x 2 where [ ] denotes the greatest 2 integer function. Q.12. Q.14 Q.16 Q.17 Q.18. 1 − tan x 1 − 2 sin x. Lim x → π4. Lim θ→ π4. 2 − cos θ − sin θ ( 4θ − π) 2. Q.15. x →0. 8 x8. x2 x2 x2 x2 1 − cos − cos + cos cos 2 4 2 4 . − cos x −1 Lim 2 x → π2 π x (x − 2 ). If Lim a sin x −3 sin 2x is finite then find the value of 'a' & the limit. x →0. tan x. x 2x a tan −1 2 , where a∈ R ; (b) Plot the graph of the function f(x) = Lim 2 t →0 π x →0 t x (ln (1 + x ) − ln 2)(3.4 x −1 − 3x ) 2 Lim 1 1 Lim [ln (1 + sin²x). cot(ln (1 + x))] Q.19 x →1 x →0 [(7 + x ) 3 − (1 + 3x ) 2 ]. sin( x − 1). −1 (a) Lim tan. n. Q.20. Q.13. Lim. π. π. . ∑ (r + 1) sin r + 1 − r sin r then find { l }. (where { } denotes the fractional part function) n →∞. If l = Lim. r =2. Q.21 Q.23 Q.24. Lim. (3x 4 + 2x 2 ) sin 1x + | x |3 +5. x →−∞. | x |3 + | x |2 + | x | +1. (x 3 + 27 ) 1n (x − 2) Q.22 Lim 2 x →3 x −9. x x x Lim 27 − 9 − 3 + 1. x →0. 2 − 1 + cos x. x , x>0 sin x = 2 − x, x ≤ 0. Let f ( x ) =. and. g( x ) = x + 3,. x <1. = x 2 − 2x − 2, 1 ≤ x < 2. = x − 5, x≥2 find LHL and RHL of g( f ( x ) ) at x = 0 and hence find Lim g( f ( x ) ) . x→0. Q.25 Q.26 Q.27. Let Pn = a. Pn −1. − 1 , ∀ n = 2, 3,.......and Let P1. = ax – 1 where a ∈ R+ then evaluate. Lim x→0. Pn x. .. 1 1 1 + ax 1 2 3 − exists and has the value equal to l, then find the value of − + . 3 x 1 + x 1 + bx a l b Let {an}, {bn}, {cn} be sequences such that (i) an + bn + cn = 2n + 1 ; (ii) anbn +bncn + cnan = 2n – 1 ; (iii) anbncn = – 1 ; (iv) an < bn < cn If the Lim x →0. Then find the value of Lim na n . n →∞. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 4 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(5) Q.28. If n ∈ N and an = 22 + 42 + 62 + ....... + (2n)2 and bn = 12 + 32 + 52 + ..... + (2n – 1)2. Find the value. Lim. a n − bn. . n Q.29 At the end points A, B of the fixed segment of length L, lines are drawn meeting in C and making angles θ and 2θ respectively with the given segment. Let D be the foot of the altitude CD and let x represents n →∞. the length of AD. Find the value of x as θ tends to zero i.e. Lim x . θ→0. Q.30. At the end-points and the midpoint of a circular arc AB tangent lines are drawn, and the points A and B are joined with a chord. Prove that the ratio of the areas of the two triangles thus formed tends to 4 as the arc AB decreases indefinitely.. EXERCISE–II 2 Q.1 Lim 2 x + 3 x →∞ 2 2x + 5 . 8 x 2 +3. x. x+c Q.2 Lim = 4 then find c x →∞ x − c 2 n 2 + n −1. Q.4. n2 + n −1 Lim n →∞ n . Q.6. a Lim cos 2 π x x →∞ 1 + x . Q.8. x Lim x − 1 + cos x x →0 x . Q.10. Let f(x) =. Q.3. F G H. x 2 sin ln cos Q.5 Lim x→∞ x2. a∈R. 1. Q.9. sin −1 (1 − {x}).cos −1 (1 − {x}) 2{x} . (1 − {x}). Q.7. Lim tan πx x →1 4 . 1 1 1 1x x x x Lim a 1 + a 2 + a 3 +.....+ a n x →∞ n . π x. (1 + x)1/ x Lim e x →0 . 1/ x. I JK. tan π2x. nx. where a1,a2,a3,......an > 0. then find xLim f(x) and xLim f(x), where {x} denotes the fractional →0 + →0 −. part function. Q.11 Q.12 Q.13 Q.14. ae x − b cos x + ce − x Find the values of a, b & c so that Lim =2 x →0 x. sin x a2 + x2 1 aπ πx Lim 2 − 2 sin sin where a is an odd integer 2 2 x →a (a − x ) 2 2 ax 2 2 Lim tan x − x. x →0. x 2 tan 2 x. If L = Lim x →1. n. Q.15. (1 − x )(1 − x 2 )(1 − x 3 )......(1 − x 2n ) [(1 − x )(1 − x 2 )(1 − x 3 ).........(1 − x n )]2. n+r r. then show that L can be equal to. 1 n ∏ ( 4 r − 2) n! r =1. (a). ∏. (c) (d). The sum of the coefficients of two middle terms in the expansion of (1 + x)2n – 1. The coefficient of xn in the expansion of (1 + x)2n.. r =1. (b). Lim [1.x ] + [ 2 .x ] + [ 3 .x ] + ..... + [ n .x ] , Where [.] denotes the greatest integer function. n →∞ n2. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 5 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(6) 1 − x + ln x 1 + cos πx. Q.16. Evaluate, Lim. Q.17. by ay exp x ln(1 + ) − exp x1n(1 + ) x x Lim Limit y y →0 x→∞ . Q.18. Let x0 = 2 cos. Q.19. 1+ x Lim ln (1 + x) − 1 2 x →0. x →1. . π and xn = 6. 2( n +1) · 2 − x n . 2 + x n −1 , n = 1, 2, 3, .........., find Lim n →∞. x. x. ∞. ∞. 4 Q.20 Let L = ∏ 1 − 2 ; M = n n =3 –1 –1 –1 L +M +N . Q.21. ∏ n =2. n3 −1 n 3 + 1 and N = . ∞. ∏ n =1. A circular arc of radius 1 subtends an angle of x radians, 0 < x <. (1 + n −1 ) 2 , then find the value of 1 + 2n −1. π as shown in 2. the figure. The point C is the intersection of the two tangent lines at A & B. Let T(x) be the area of triangle ABC & let S(x) be the area of the shaded region. Compute: (a) T(x). (b) S(x) n. Q.22. x. 3n −1 sin 3 n ∑ n →∞ 3. Let f (x) = Lim. &. (c) the limit of. and g (x) = x – 4 f (x). Evaluate Lim (1 + g( x ) )cot x . x →0. n =1. Q.23. Q.24. T (x) as x → 0. S(x). n 2 θ If f (n, θ)= ∏ 1 − tan r , then compute Lim f (n, θ) 2 n →∞ r =1 . L = Lim x →0. cos 2 x + (1 + 3x )1 3 3 4 cos3 x − ln (1 + x ) 4 − 2 4 x. If L = a b where 'a' and 'b' are relatively primes find (a + b). x2. Q.25. cosh (π x ) Lim x → ∞ cos (π x ) . Q.26. f (x) is the function such that Lim. where cosh t =. x →0. et + e−t 2. f (x ) x (1 + a cos x ) − b sin x = 1 . If Lim = 1 , then find the value of x x →0 (f (x ) )3. a and b. Q.27. Through a point A on a circle, a chord AP is drawn & on the tangent at A a point T is taken such that AT = AP. If TP produced meet the diameter through A at Q, prove that the limiting value of AQ when P moves upto A is double the diameter of the circle.. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 6 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(7) Q.28. Using Sandwich theorem, evaluate (a). 1 1 1 1 Lim + + + ........... + 2 2 2 2 n →∞ n +1 n +2 n + 2n n. (b). Lim n →∞. 1 2 n + + ......... + 1+ n2 2 + n 2 n + n2. Q.29. x2 + 1 Lim − ax − b = 0 Find a & b if : (i) x →∞ . Q.30. 1 1 then find the value of L + 153 . If L = Lim − L x →0 ln (1 + x ) ln ( x + 1 + x 2 ) . Q.1. x Lim x tan 2x − 2x tan is : 2 x →0 (1 − cos 2x). x +1. . x 2 − x + 1 − ax − b = 0 (ii) xLim →−∞ . EXERCISE–III. (A) 2. (B) − 2. [ JEE '99, 2 (out of 200) ] (C). x. Q.2. x − 3 For x ∈ R , Lim = x →∞ x + 2. (A) e Q.3. (B) e −1. (B) π. Q.4. a tan x − a sin x Evaluate Lim , a > 0. x →0 tan x − sin x. Q.5. The integer n for which Lim (A) 1 If Lim x →0. (A). Q.7. 1 n. (D) −. 1 2. [ JEE 2000, Screening] (C) e −5. sin( π cos 2 x ) Lim equals x →0 x2. (A) –π. Q.6. 1 2. (D) e5 [ JEE 2001, Screening]. (C). π 2. (D) 1 [REE 2001, 3 out of 100]. (cos x − 1)(cos x − e x ) is a finite non-zero number is x →0 xn (B) 2 (C) 3 (D) 4 [JEE 2002 (screening), 3] sin(n x )[(a − n )n x − tan x ] = 0 (n > 0) then the value of 'a' is equal to x2 n2 +1 (B) n2 + 1 (C) (D) None n [JEE 2003 (screening)]. 2 1 (n + 1) cos −1 − n . Find the value of Lim n →∞ π n . [ JEE ' 2004, 2 out of 60]. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 7 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(8) KEY CONCEPTS (CONTINUITY) THINGS TO REMEMBER : 1. A function f(x) is said to be continuous at x = c, if Limit f(x) = f(c). Symbolically x →c f is continuous at x = c if Limit f(c - h) = Limit f(c+h) = f(c). h →0. h →0. i.e. LHL at x = c = RHL at x = c equals Value of ‘f’ at x = c. It should be noted that continuity of a function at x = a is meaningful only if the function is defined in the immediate neighbourhood of x = a, not necessarily at x = a. 2. (i). Reasons of discontinuity: Limit f(x) does not exist x →c. i.e. Limit− f(x) ≠ Limit f (x) + x →c. (ii) (iii). x →c. f(x) is not defined at x= c Limit f(x) ≠ f (c) x →c. Geometrically, the graph of the function will exhibit a break at x= c. The graph as shown is discontinuous at x = 1 , 2 and 3. 3. Types of Discontinuities : Type - 1: ( Removable type of discontinuities) In case Limit f(x) exists but is not equal to f(c) then the function is said to have a removable discontinuity x →c. or discontinuity of the first kind. In this case we can redefine the function such that Limit f(x) = f(c) & x →c. make it continuous at x= c. Removable type of discontinuity can be further classified as : (a). MISSING POINT DISCONTINUITY : Where Limit f(x) exists finitely but f(a) is not defined. x →a. sin x (1 − x )(9 − x 2 ) e.g. f(x) = has a missing point discontinuity at x = 1 , and f(x) = has a missing point x (1 − x ) discontinuity at x = 0. (b). ISOLATED POINT DISCONTINUITY : Where Limit f(x) exists & f(a) also exists but ; Limit ≠ f(a). x →a. x 2 − 16 e.g. f(x) = , x ≠ 4 & f (4) = 9 has an isolated point discontinuity at x = 4. x−4. x →a. 0 if x ∈ I Similarly f(x) = [x] + [ –x] = has an isolated point discontinuity at all x ∈ I. −1 if x ∉ I. Type-2: ( Non - Removable type of discontinuities) In case Limit f(x) does not exist then it is not possible to make the function continuous by redefining it. x →c. Such discontinuities are known as non - removable discontinuity or discontinuity of the 2nd kind. Non-removable type of discontinuity can be further classified as : (a). Finite discontinuity e.g. f(x) = x − [x] at all integral x ; f(x) = tan −1 ( note that f(0+) = 0 ; f(0–) = 1 ). (b). (c). 1 at x = 0 and f(x) = x. 1 1 1+ 2 x. at x = 0. π cosx 1 tanx at x = Infinite discontinuity e.g. f(x) = 1 or g(x) = at x = 4 ; f(x) = 2 and f(x) = 2 x ( x − 4) 2 x−4 at x = 0. Oscillatory discontinuity e.g. f(x) = sin 1 at x = 0. x ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 8 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(9) In all these cases the value of f(a) of the function at x= a (point of discontinuity) may or may not exist but Limit does not exist. x →a. Note: From the adjacent graph note that – f is continuous at x = – 1 – f has isolated discontinuity at x = 1 – f has missing point discontinuity at x = 2 – f has non removable (finite type) discontinuity at the origin. 4.. In case of dis-continuity of the second kind the non-negative difference between the value of the RHL at x = c & LHL at x = c 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.. 5.. All Polynomials, Trigonometrical functions, exponential & Logarithmic functions are continuous in their domains.. 6.. If f & g are two functions that are continuous at x= c then the functions defined by : F1(x) = f(x) ± g(x); F2(x) = K f(x), K any real number; F3(x) = f(x).g(x) are also continuous at x= c. Further, if g (c) is not zero, then F4(x) =. 7.. f (x) is also continuous at x= c. g(x). The intermediate value theorem: Suppose f(x) is continuous on an interval I , and a and b are any two points of I. Then if y0 is a number between f(a) and f(b) , their exists a number c between a and b such that f(c) = y0.. The function f, being continuous on [a,b) takes on every value between f(a) and f(b). NOTE VERY CAREFULLY THAT : (a) If f(x) is continuous & g(x) is discontinuous at x = a then the product function φ(x) = f(x). g(x) is not necessarily be discontinuous at x = a. e.g. sin πx. f(x) = x & g(x) = . 0. (b). x≠0 x=0. If f(x) and g(x) both are discontinuous at x = a then the product function φ(x) = f(x). g(x) is not necessarily be discontinuous at x = a. e.g. 1. f(x) = − g(x) = −1. x≥0 x<0. (c). Point functions are to be treated as discontinuous. eg. f(x) = 1 − x +. (d). A Continuous function whose domain is closed must have a range also in closed interval.. (e). If f is continuous at x = c & g is continuous at x = f(c) then the composite g[f(x)] is continuous at x = c. eg. f(x) =. x sin x x2 + 2. be continuous at. x −1 is not continuous at x = 1.. & g(x) = x are continuous at x = 0 , hence the composite (gof) (x) =. x sin x x2 + 2. will also. x=0.. 7.. CONTINUITY IN AN INTERVAL :. (a). A function f is said to be continuous in (a , b) if f is continuous at each & every point ∈(a , b). ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 9 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(10) (b) (i) (ii). A function f is said to be continuous in a closed interval [a ,b ] if : f is continuous in the open interval (a , b) & f is right continuous at ‘a’ i.e. Limit+ f(x) = f(a) = a finite quantity.. (iii). f is left continuous at ‘b’ i.e. Limit− f(x) = f(b) = a finite quantity.. x →a. x→b. Note that a function f which is continuous in [a ,b ] possesses the following properties :. (i). If f(a) & f(b) possess opposite signs, then there exists at least one solution of the equation f(x) = 0 in the open interval (a , b).. (ii). If K is any real number between f(a) & f(b), then there exists at least one solution of the equation f(x) = K in the open inetrval (a , b).. 8.. SINGLE POINT CONTINUITY: Functions which are continuous only at one point are said to exhibit single point continuity e.g. f(x) =. x if x ∈Q x if x ∈ Q and g(x) = are both continuous only at x = 0. 0 if x ∉Q − x if x ∉ Q. EXERCISE–I Q.1. 3x 2 + ax + a + 3 If the function f (x) = is continuous at x = – 2. Find f (–2). x2 + x − 2. Q.2. Find all possible values of a and b so that f (x) is continuous for all x ∈ R if. | ax + 3 | | 3x + a | f (x) = b sin 2x − 2b x cos 2 x − 3. if x ≤ −1 if − 1 < x ≤ 0 if 0 < x < π if x ≥ π. ln cos x. if x > 0 2 + − 1 x 1 Let f(x) = esin 4 x − 1 if x < 0 ln (1 + tan 2x ) 4. Q.3. Is it possible to define f(0) to make the function continuous at x = 0. If yes what is the value of f(0), if not then indicate the nature of discontinuity. Q.4. f (x) x−3 K. Suppose that f (x) = x3 – 3x2 – 4x + 12 and h(x) =. , x≠3 , x=3. then. (a) find all zeros of f (x) (b) find the value of K that makes h continuous at x = 3 (c) using the value of K found in (b), determine whether h is an even function. Q.5. Q.6. x2 x2 x2 + + ............ + and y (x) = Lim y n ( x ) Let yn(x) = + 1 + x 2 (1 + x 2 )2 (1 + x 2 ) n −1 n →∞ Discuss the continuity of yn(x) (n ∈ N) and y(x) at x = 0 x2. Draw the graph of the function f(x) = x −x − x², −1 ≤ x ≤ 1 & discuss the continuity or discontinuity of f in the interval −1 ≤ x ≤ 1.. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 10 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(11) Q.7. Let f(x) = . 1−sin πx 1+ cos 2 πx. ,. p, 2 x −1 4+ 2 x −1 − 2. x < 12 x = 12 . Determine the value of p, if possible, so that the function is continuous ,x>. 1 2. at x=1/2. Q.8. Given the function g (x) = 6 − 2 x and h (x) = 2x2 – 3x + a. Then g( x ), x ≤ 1 (a) evaluate h ( g(2) ) (b) If f (x) = , find 'a' so that f is continuous. h ( x ), x > 1. Q.9. . Determine the form of g(x) = f [f(x)] & hence find the point of Let f(x) = 3 − x , 2 < x ≤ 3 discontinuity of g , if any.. Q.10. Let [x] denote the greatest integer function & f(x) be defined in a neighbourhood of 2 by. 1 + x , 0 ≤ x ≤ 2. [ x +1]. 4 −16 (exp {( x +2) ln4}) , x <2 x 4 −16 f (x) = . 1−cos( x −2) , x >2 A ( x −2) tan ( x −2) Find the values of A & f(2) in order that f(x) may be continuous at x = 2.. (). tan6x. Q.11. 6 tan5x 5 The function f(x) = b+2 a tan x (1+ cosx ) b . if if. 0<x< π2 x= π2. if. π <x<π 2. Determine the values of 'a' & 'b' , if f is continuous at x = π/2. Q.12. ax 2 + bx + c + e nx A function f : R → R is defined as f (x) = Lim where f is continuous on R. Find the n →∞ 1 + c ·e nx. values of a, b and c. Q.13. Q.14. (x − 1)1/ 3 , x < 0 . Discuss the continuity of g (f (x)). 1/ 2 , x≥0 (x + 1) 1−sin 3 x if x < π 2 3cos 2 x π π if x = 2 Determine a & b so that f is continuous at x = . f(x) = a 2 b (1− sin x ) if x > π2 2 (π − 2 x ). 1 + x 3 , x < 0 ; g(x) = x 2 − 1 , x ≥ 0. Let f(x) = . Q.15 Determine the values of a, b & c for which the function f (x) = is continuous at x = 0.. sin(a +1)x+sinx x. c. (x+bx ). 2 1/ 2. −x1/ 2. bx3 / 2. for x<0 for x=0 for x>0. Q.16. If f(x) = sin 3 x + A sin 2 x + B sin x (x≠ 0) is cont. at x = 0. Find A & B. Also find f(0).. Q.17. 1 x 3−1 Discuss the continuity of the function ‘f’ defined as follows: f(x) = x + 1 x +1 x − 5. x5. for 0 ≤ x ≤ 2 for 2 < x ≤ 4 and draw the for 4 < x ≤ 6. graph of the function for x ∈ [0, 6]. Also indicate the nature of discontinuities if any.. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 11 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(12) Q.18 Q.19. Q.20. If f(x) = x + {-x} + [x] , where [x] is the integral part & {x} is the fractional part of x. Discuss the continuity of f in [ – 2, 2 ]. ax − b for x ≤1 Find the locus of (a, b) for which the function f (x) = 3x for 1 < x < 2 2 is continuous at x = 1 but discontinuous at x = 2. x≥2 bx − a for. sin 2 (π·2 x ) x n f (x ) + h(x) + 1 Lim , x ≠ 1 and g (1) = x →1 be a continuous function Let g (x) = Lim ln sec(π·2 x ) n →∞ 2 x n + 3x + 3 at x = 1, find the value of 4 g (1) + 2 f (1) – h (1). Assume that f (x) and h (x) are continuous at x = 1.. (. ). Q.21. If g : [a, b] onto [a, b] is continous show that there is some c ∈ [a, b] such that g (c) = c.. Q.22. 2 + cos x 3 − is not defined at x = 0. How should the function be defined at The function f(x) = 3 x sin x x 4 x = 0 to make it continuous at x = 0.. Q.23. f (x) =. a sin x − a tan x for x > 0 tan x − sin x. ln (1 + x + x 2 ) + ln (1 − x + x 2 ) for x < 0, if f is continuous at x = 0, find 'a' sec x − cos x x now if g (x) = ln 2 − · cot (x – a) for x ≠ a, a ≠ 0, a > 0. If g is continuous at x = a then show that a –1 g(e ) = – e. =. Q.24(a) Let f(x + y) = f(x) + f(y) for all x , y & if the function f(x) is continuous at x = 0 , then show that f(x) is continuous at all x. (b) If f(x. y) = f(x). f(y) for all x , y and f(x) is continuous at x = 1. Prove that f(x) is continuous for all x except at x = 0. Given f(1) ≠ 0. n. Q.25. Given f (x) =. ∑ tan xr sec . 2 . r=1. (. x . 2. r − 1 . ) ( (. ; r, n ∈ N. ) [ ( )] n. l n f (x ) + tan xn − f (x) + tan xn . sin tan x 2 2 2 Limit g (x) = n → ∞ n 1 + f (x) + tan xn 2 π = k for x = and the domain of g (x) is (0 , π/2). 4. Q.26. ). where [ ] denotes the greatest integer function. Find the value of k, if possible, so that g (x) is continuous at x = π/4. Also state the points of discontinuity of g (x) in (0 , π/4) , if any. f (x ) Let f (x) = x3 – x2 – 3x – 1 and h (x) = where h is a rational function such that g( x ) 1 (a) it is continuous every where except when x = – 1, (b) Lim h ( x ) = ∞ and (c) Lim h ( x ) = . x →−1 2 x →∞ Find Lim (3h ( x ) + f ( x ) − 2g ( x ) ) x →0. Q.27. 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 1 0, 2 such that f (c) = f c + 2 . ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 12 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(13) Q.28. 1 − ax + xax l na for x < 0 x 2 a x where a > 0. Consider the function g(x) = x x 2 a − xln2 − xl na − 1 for x > 0 x2 . find the value of 'a' & 'g(0)' so that the function g(x) is continuous at x = 0.. (. Q.29. (. )). π −sin −1 1−{x}2 .sin −1(1−{x}) for x ≠0 2 3 − 2 { x } { x } Let f(x) = where {x} is the fractional part of x. π for x =0 2 . (. ). Consider another function g(x) ; such that g(x) = f(x) for x ≥ 0 = 2 2 f(x) for x < 0 Discuss the continuity of the functions f(x) & g(x) at x = 0. Q.30. 4 x − 5 [x] for x > 1 ; where [x] is the greatest for x ≤ 1 [ cos π x]. Discuss the continuity of f in [0,2] where f(x) = integer not greater than x. Also draw the graph.. EXERCISE–II (OBJECTIVE QUESTIONS) Q.1. State whether True or False.. (i). f(x) = Limit n →∞. (ii). The function defined by f(x)=. (iii). The function f(x) = 2 − 2. (iv). There exists a continuous function f: [0, 1] onto [0, 10], but there exists no continuous function g : [0, 1] onto (0, 10).. (v). If f (x) is continuous in [0 , 1] & f(x) = 1 for all rational numbers in [0 , 1] then f 1. (vi). cos πx + sin( πx 2) if x ≠ 1 2 3π 2 ( x − 1 )( 3 x − 2 x − 1 ) If f (x) = is continuous, then the value of k is . 32 k if k = 1 Select the correct alternative : (Only one is correct). Q.2. f is a continuous function on the real line. Given that. 1 1 + n sin 2 π x. is continuous at x = 1.. 1 /(1− x ). x for x ≠ 0 & f(0) = 1 is continuous at x = 0. | x | +2 x 2. if x ≠ 1 & f(1) = 1 is not continuous at x = 1.. (. x² + (f(x) − 2) x − 3 . f(x) + 2 3 − 3 = 0. Then the value of f( 3 ) (A) can not be determined Q.3. (B) is 2 (1 − 3 ). (C) is zero. 2. (D) is. (. ). 2 equal to 1.. ). 3−2. If f (x) = sgn (cos 2 x − 2 sin x + 3) , where sgn ( ) is the signum function , then f (x) (A) is continuous over its domain (B) has a missing point discontinuity (C) has isolated point discontinuity (D) has irremovable discontinuity.. 3. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 13 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(14) Q.4. Q.5. [x ] {x}, h(x) = g(f (x)) where {x} denotes fractional part and [x + 1] [x] denotes the integral part then which of the following holds good? (A) h is continuous at x = 0 (B) h is discontinuous at x = 0 (C) h(0–) = π/2 (D) h(0+) = – π/2. Let g(x) = tan–1|x| – cot–1|x|, f(x) =. x n − sin x n Consider f (x) = Limit n for x > 0, x ≠ 1, n → ∞ x + sin x n f (1) = 0 then (A) f is continuous at x = 1 (B) f has an infinite or oscillatory discontinuity at x = 1. (C) f has a finite discontinuity at x = 1 (D) f has a removable type of discontinuity at x = 1.. [{| x |}]e x {[x + {x}]} 2. Q.6. Given f(x) =. e1 x 2 − 1 sgn (sin x ) . for x ≠ 0. = 0 for x = 0 where {x} is the fractional part function; [x] is the step up function and sgn(x) is the signum function of x then, f(x) (A) is continuous at x = 0 (B) is discontinuous at x = 0 (C) has a removable discontinuity at x = 0 (D) has an irremovable discontinuity at x = 0. Q.7. Q.8. x[ x ]2 log (1+ x ) 2 for− 1 < x < 0 Consider f(x) = ln e x 2 + 2 {x} for0 < x < 1 tan x where [ * ] & {*} are the greatest integer function & fractional part function respectively, then (A) f(0) = ln2 ⇒ f is continuous at x = 0 (B) f(0) = 2 ⇒ f is continuous at x = 0 2 (C) f(0) = e ⇒ f is continuous at x = 0 (D) f has an irremovable discontinuity at x = 0 1+ x − 1− x , x ≠ 0; {x} π < x < 0, g (x) = cos 2x, – 4 1 f (g(x)) for x < 0 2 for x = 0 h(x) – 1 f (x) forx > 0. Consider. f (x) =. then, which of the following holds good. where {x} denotes fractional part function. (A) 'h' is continuous at x = 0 (C) f(g(x)) is an even function. (B) 'h' is discontinuous at x = 0 (D) f(x) is an even function. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 14 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(15) Q.9. The function f(x) = [x]. cos (A) all x (C) no x. Q.10. 2x − 1 π , where [•] denotes the greatest integer function, is discontinuous at 2 (B) all integer points (D) x which is not an integer. Consider the function defined on [0, 1] → R, f (x) =. sin x − x cos x if x ≠ 0 and f (0) = 0, then the x2. function f (x) (A) has a removable discontinuity at x = 0 (B) has a non removable finite discontinuity at x = 0 (C) has a non removable infinite discontinuity at x = 0 (D) is continuous at x = 0. Q.11. a − x π x sin tan for x > a 2 2a f (x) = cos π x 2a for x < a a−x. where [x] is the greatest integer function of x, and a > 0, then (A) f (a–) < 0 (B) f has a removable discontinuity at x = a (C) f has an irremovable discontinuity at x=a (D) f (a+) < 0. sin πx − x 2 n sin( x − 1) Q.12 Consider the function f (x) = Lim , where n ∈ N n →∞ 1 + x 2n +1 − x 2 n Statement-1: f (x) is discontinuous at x = 1. because Statement-2: f (1) = 0. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. Q.13. Consider the functions f (x) = sgn (x – 1) and g (x) = cot–1[x – 1] where [ ] denotes the greatest integer function. Statement-1 : The function F (x) = f (x) · g (x) is discontinuous at x = 1. because Statement-2 : If f (x) is discontinuous at x = a and g (x) is also discontinuous at x = a then the product function f (x) · g (x) is discontinuous at x = a. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 15 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(16) Select the correct alternative : (More than one are correct) Q.14. A function f is defined on an interval [a, b]. Which of the following statement(s) is/are INCORRECT? (A) If f (a) and f (b), have opposite sign, then there must be a point c ∈ (a, b) such that f (c) = 0. (B) If f is continuous on [a, b], f (a) < 0 and f (b) > 0, then there must be a point c ∈ (a, b) such that f (c) = 0 (C) If f is continuous on [a, b] and there is a point c in (a, b) such that f (c) = 0, then f (a) and f (b) have opposite sign. (D) If f has no zeroes on [a, b], then f (a) and f (b) have the same sign.. Q.15. Which of the following functions f has/have a removable discontinuity at the indicated point? (A) f (x) =. x 2 − 2x − 8 at x = – 2 x+2. x 3 + 64 (C) f (x) = at x = – 4 x+4. Q.16. (B) f (x) =. x −7 at x = 7 | x −7|. (D) f (x) =. 3− x at x = 9 9−x. (. ). 2. n −n Let ‘f’ be a continuous function on R. If f (1/4n) = sin e e +. (A) not unique (C) data sufficient to find f(0). n2 then f(0) is : n2 +1. (B) 1 (D) data insufficient to find f(0). x − 1 , then on the interval [0, π] 2 1 1 are both continuous (B) tan ( f (x) ) & are both discontinuous (A) tan ( f (x) ) & f (x ) f (x ) 1 (C) tan ( f (x) ) & f −1 (x) are both continuous (D) tan ( f (x) ) is continuous but is not f (x ). Q.17. Indicate all correct alternatives if, f (x) =. Q.1. The function f(x) = [x]2 − [x2] (where [y] is the greatest integer less than or equal to y), is discontinuous. EXERCISE–III at : (A) all integers (C) all integers except 0. (B) all integers except 0 & 1 (D) all integers except 1 [ JEE '99, 2 (out of 200) ]. Q.2. 1/ x for x < 0 (1 + ax) Determine the constants a, b & c for which the function f(x) = b for x = 0 is continuous at (x + c)1/3 − 1 for x > 0 [ REE '99, 6 ] x = 0. 1/2 (x + 1) − 1. Q.3. Discuss the continuity of the function. e1/(x −1) − 2 , 1/( x −1) +2 f(x) = e 1, at x = 1.. x≠1 x =1 [REE 2001 (Mains) , 3 out of 100]. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 16 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(17) ANSWER KEY LIMIT EXERCISE–I Q 1. 3. Q 2.. 45 91. 3 2. Q.8. p−q 2. Q.7. Q 3. 2 Q.9 a =. Q.4. 1 1 2 ;r= ;S= 2 4 3. Q.11 (a) does not exist; (b) does not exist; (c) 0. 5050. Q 5.. 2 3. Q.6. –. Q 10. l n 2. Q 12. 2. Q.13. 1 32. Q.14. 1 16 2. Q.15 21n 2 π. Q.16 a = 2 ; limit = 1 Q.17 (a) π/2 if a > 0 ; 0 if a = 0 and –π/2 if a < 0; (b) f(x) = | x |. Q.18 1. Q 19. −. 9 4 1n 4 e. Q.20 π – 3 Q.21 −2. Q.25 (ln a)n. Q.24 – 3, –3, – 3. Q.26 72. 1 3. Q.23 8 2 (1n 3) 2. Q.22 9. Q.27 – 1/2 Q.28. 3 2. Q.29. 2L Q.30 4 3. EXERCISE–II Q.1. e-8. Q.2 c = ln2. Q.3 e. − 12. Q.4. e–1. Q.5. π2 − 4. Q.6 e − 2 π. 2 a2. Q.7. e-1 Q.8 e–1/2. Q.9 (a1.a2.a3....an). Q.10. π π , 2 2 2. Q.11 a = c = 1, b = 2. π 2a 2 + 4 Q.12 16a 4. Q.13. 2 3. Q.15. x 2. Q.16 –. Q.21 T(x) =. 1 2. x 2. 1 π2. tan2 . sin x or tan. Q.17 a - b. Q.18. π 3. Q.19 1/2. Q.20 8. 3 x sin x 1 1 − , S(x) = x − sin x, limit = 2 2 2 2 2. θ tan θ. π Q.25 e. Q.24 19. 2. Q.22 g (x) = sin x and l = e. Q.23. Q.26 a = – 5/2, b = – 3/2. Q.28 (a) 2; (b) 1/2 Q.29 (i) a =1, b = −1 (ii) a = −1 , b =. 1 2. Q.30 307. EXERCISE–III Q.1. C. Q.2. C. Q.6. C. Q.7. 1−. Q.3. B. Q.4. lna. Q.5. C. 2 π. *************************. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 17 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
(18) CONTINUITY EXERCISE–I Q.1 Q.3 Q.4 Q.5 Q.6. –1 Q.2 a = 0, b = 1 + – f(0 ) = –2 ; f(0 ) = 2 hence f(0) not possible to define (a) −2, 2, 3 (b) K = 5 (c) even yn(x) is continuous at x = 0 for all n and y(x) is dicontinuous at x = 0 f is cont. in −1 ≤ x ≤ 1 Q.7 P not possible.. (a) 4 – 3 2 + a , (b) a = 3 g(x) = 2 + x for 0 ≤ x ≤ 1, 2 − x for 1 < x ≤ 2, 4 − x for 2 < x ≤ 3 , g is discontinuous at x = 1 & x = 2 A = 1 ; f(2) = 1/2 Q.11 a = 0 ; b = −1 Q.12 c = 1, a, b ∈ R gof is dis-cont. at x = 0, 1 & – 1 a = 1/2, b = 4 Q.15 a = − 3/2, b ≠ 0, c = 1/2 A = − 4 , B = 5, f(0) = 1 Q.17 discontinuous at x = 1, 4 & 5 discontinuous at all integral values in [− 2 , 2] locus (a, b) → x, y is y = x – 3 excluding the points where y = 3 intersects it. 1 Q.20 5 Q.22 60. Q.8 Q.9 Q.10 Q.13 Q.14 Q.16 Q.18 Q.19. l n (tan x). Q.25 k = 0 ; g (x) = . 0. if 0 < x < π 4 . Hence g (x) is continuous everywhere. if π ≤ x < π 4 2. Q.26 g (x) = 4 (x + 1) and limit = – Q.28 a = Q.29. 1 2. 39 4. 2 l n 2) ( , g(0) =. 8. π π f(0+) = ; f(0−) = ⇒ f is discont. at x = 0 ; 2 4 2. g(0+) = g(0−) = g(0) = π/2 ⇒ g is cont. at x = 0. Q.30 the function f is continuous everywhere in [0 , 2] except for x = 0 , 1 , 1 & 2. 2. EXERCISE–II Q.1 Q.2 Q.9 Q.15. (i) false ; (ii) false ; (iii) true ; (iv) true; (v) true; (vi) True B Q.3 C Q.4 A Q.5 C Q.6 A C Q.10 D; Q.11 B Q.12 B Q.13 C A, C, D Q.16 B, C Q.17 C, D. Q.7 D Q.8 Q.14 A, C, D. A. EXERCISE–III D. Q.3. Discontinuous at x = 1; f(1+) = 1 and f(1–) = –1 *************************. Q.2. a = ln. 2 2 ; b= ; c=1 3 3. Q.1. ETOOS Academy Pvt. Ltd. : F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors 18 (Mahindra Showroom), BSNL Office Lane, Jhalawar Road, Kota, Rajasthan (324005).
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