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MANOJ CHAUHAN SIR(IIT-DELHI) EX. SR. FACULTY (BANSAL CLASSES)

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)

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

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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 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

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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.

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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]

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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)

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 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.

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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 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 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

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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

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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

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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.

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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]

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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

*************************

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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

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MANOJ CHAUHAN SIR(IIT-DELHI) EX. SR. FACULTY (BANSAL CLASSES)

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)

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

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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 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

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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.

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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]

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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)

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 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.

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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 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 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

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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

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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

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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.

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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]

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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

*************************

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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

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