Limit Continuity

Example : 1 Find the domain and the range of the following functions (a) Solution (a)

y=

(b)

1− x 2

(c)

(c)

y=

1 x−2

1 – x2 ≥ 0 ⇒ x2 ≤ 1 ⇒ –1≤x≤1 Hence the domain is x set [–1, 1]. For range : As –1≤x≤1 ⇒ 0 ≤ x2 ≤ 1 ⇒ 0 ≤ 1 – x2 ≤ 1

For domain :

⇒

(b)

y = 2 sin x

0≤

1− x 2 ≤ 1 ⇒ 0≤y≤1 Hence the range is set [0, 1] y = 2 sin x For domain : x ∈ R i.e. x (–∞, ∞) For range : – 1 ≤ sin x ≤ 1 – 2 ≤ 2 sin x ≤ 2 –2≤y≤2 Hence the range is y ∈ [–2, 2] As denominator cannot be zero, x can not be equal to 2 domain is x ∈ R – {2} i.e. x ∈ (–∞ , 2) (2, ∞) Range : As y can never become zero, the range is y ∈ R – {0} i.e. y ∈ (–∞, 0) (0, ∞)

Example : 2 Find the domain of the following functions : (a)

1 3 − x + log x 10

(b)

1 x+ | x |

(c)

1 − log10 x

Solution (a)

3 − x is defined if 3 – x ≥ 0 ⇒

x≤3

............(i)

1 log10 x is defined if x > 0 and x ≠ 1

(b)

(c)

⇒ x > 0 – {1} ............(ii) Combining (i) and (ii), set of domain is : x ∈ (0, 1) ∪ (1, 3] f(x) is defined if : x + |x| ≠ 0 ⇒ |x| ≠ – x ⇒ x>0 Hence domain is x ∈ (0, ∞) f(x) is defined if 1 – log10x ≥ 0 and x>0 ⇒ log10x ≤ 1 and x>0 ⇒ x ≤ 10 and x>0 ⇒ domain is x ∈ (0, 10]

Example : 3 (Using factorisation) Evaluate the following limits : (a)

3 lim x − 2 x − 4 x →2 x 2 − 3 x + 2

(b)

3 3 lim x − a x →a x 2 − ax

(c)

4 lim x − 625 x →5 x 3 − 125

Page # 1.

Solution (a)

2 3 2 lim x − 2 x − 4 = lim ( x − 2)( x + 2x + 2) = lim x + 2x + 2 = 10 x →2 x 2 − 3 x + 2 x →2 x →2 ( x − 2)( x − 1) x −1

(b)

2 2 3 3 2 2 lim x − a = lim ( x − a)( x + ax + a ) = lim x + ax + a = 3a x →a x 2 − ax x →a x →a x( x − a ) x

(c)

x 4 − 54 4 4 .5 3 20 x−5 lim x − 625 = lim = = 3 3 3 x →5 x − 125 x →5 x − 5 3 3 .5 2 x−5

[using section 2.2 (ix)]

Example : 4 (Using rationalisation) Evaluate the following limits : (a)

3x + 7 − 4

lim

x →3

(b)

x +1− 2

lim

x →a

a + 2x − 3 x 3a + x − 2 x

(c)

lim

x →1

3

x −1

x2 − 1

Solution (a)

3x + 7 − 4

lim

x →3

x +1− 2

Rationalising the numerator and denominator, 3 x + 7 − 16 = xlim →3 x + 1− 4 3( x − 3 ) = xlim →3 x−3

= 3 xlim →3 (b)

⎛ x +1+ 2 ⎞ ⎟ ⎜ ⎜ 3x + 7 + 4 ⎟ ⎠ ⎝

⎛2+2⎞ 3 ⎟ = =3 ⎜ 4 + 4 ⎝ ⎠ 2 3x + 7 + 4

x +1 + 2

Rationalising numerator and denominator we get, a + 2x − 3 x = xlim →a 3a + x − 4 x

= xlim →a

(c)

⎛ x +1+ 2 ⎞ ⎟ ⎜ ⎜ 3x + 7 + 4 ⎟ ⎠ ⎝

a−x 3(a − x )

⎛ 3a + x + 2 x ⎞ ⎟ ⎜ ⎜ a + 2x + 3 x ⎟ ⎠ ⎝

⎛ 3a + x + 2 x ⎞ 1 ⎟ ⎜ ⎜ a + 2x + 3 x ⎟ = 3 ⎠ ⎝

⎛2 a +2 a ⎞ 2 ⎟ ⎜ ⎜ 3a + 3a ⎟ = 3 3 ⎠ ⎝

1 ⎛ 1 ⎞⎛ 2 ⎞ ⎜ x 3 − 1⎟⎜ x 3 + x 3 + 1⎟ 1 ⎜ ⎟⎜ ⎟ ⎠⎝ ⎠ lim x 3 − 1 = lim ⎝ x →1 x →1 2 1 2 ⎛ ⎞ x −1 ( x 2 − 1)⎜ x 3 + x 3 + 1⎟ ⎜ ⎟ ⎝ ⎠

lim

x →1

( x − 1) 1 = 2 1 ⎛ ⎞ 6 ( x − 1)( x + 1)⎜ x 3 + x 3 + 1⎟ ⎜ ⎟ ⎝ ⎠

Page # 2.

Example : 5 (x → ∞ type problems) Evaluate the following limits : (a)

3 2 lim x − 2x + 3 x + 1 x →∞ 5 x 3 + 7x + 2

(b)

2 2 2 2 lim 1 + 2 + 3 + ........ + n n→ ∞ n3

(c)

lim

x →∞

x 2 + 3 x + 1 + 5x 1 + 4x

Solution In these type of problems, divide numerator and denominator by highest power of x. (a) Dividing numerator and denominator by x3

2 3 1 + 2 + 3 x x x 1 = 7 2 5 5+ 2 + 3 x x

1− = xlim →∞

[because as x → ∞,

(b)

2 2 2 2 lim 1 + 2 + 3 + ........ + n n→ ∞ n3

= nlim →∞

(c)

1 1 1 , , ......... → 0] x x2 x3

n(n + 1)(2n + 1) 6n3

=

1⎞ ⎛ 1⎞ 1 lim ⎛ ⎜1 + ⎟ ⎜ 2 + ⎟ n → ∞ n⎠ 6 ⎝ n⎠ ⎝

=

1 1 (1 + 0) (2 + 0) = 6 3

The highest power of x is 1. Hence divide the numerator and denominator by x. x 2 + 3 x + 1 + 5x 1 + 4x

lim

x →∞

1+ = xlim →∞

3 1 + +5 x x2 1+5 3 = = 1 0+4 2 +4 x

Example : 6 (∞ – ∞ form) Evaluate the following limits :

lim (a)

x→

π 2

(sec x – tan x)

(b)

lim

x →∞

( ( x + 2a)(2x + a) − x 2 )

(c)

lim ⎛⎜ x − x 2 + x ⎞⎟ ⎝ ⎠

x →∞

Solution

lim

(a)

π x→ 2

lim x→

(b)

π 2

lim

x →∞

(sec x – tan x) =

lim sec 2 x − tan 2 x π x→ 2 sec x + tan x

1 =0 sec x + tan x

( ( x + 2a)(2x + a) − x 2 )

Rationalising the expression, we get

lim

x →∞

( x + 2a)(2x + a) − 2x 2

( (x + 2a)(2x + a) + x 2 ) Page # 3.

5ax + 2a 2

= xlim →∞

2x 2 + 5ax + 2a 2 + x 2

Dividing numerator and denominator by x, we get 5a +

= xlim →∞

(c)

2+

2a 2 x

5a

2

5a 2a + 2 + 2 x x

=

2 2

lim ⎛⎜ x − x 2 + x ⎞⎟ ⎝ ⎠

x →∞

On rationalising the expression, we get x 2 − (x 2 + x)

= xlim →∞

= xlim →∞

x + x2 + x

−x x + x2 + x

Divide by the highest power of x i.e. x1

−1

= xlim →∞

=

1 1+ 1+ x

−1 1 =– 1+ 1 2

Example : 7 sin x ⎛ ⎞ = 1⎟ Evaluate the following limits : ⎜ u sin g lim x →0 x ⎝ ⎠

lim tan x − 3 π

(a)

x→

(c)

x →a

3

9x − π 2

2

lim cos x − cos a x−a

lim sin(cos x ) cos x π 2 sin x − cos ecx

(b)

x→

(d)

x →a

lim

a sin x − x sin a ax 2 − a 2 x

Solution

(a)

lim tan x − 3 π

x→

3

9x 2 − π2

Using tan A – tan B =

lim x→

(b)

π 3

π 3 9x 2 − π2

lim tan x −

=

π x→ 3

sin( A − B) we get, cos A cos B

π⎞ ⎛ sin⎜ x − ⎟ sin θ 1 ⎛ ⎞ 1 2 3 ⎝ ⎠ = 1⎟ = ⎜ u sin g lim = θ → 0 π π θ π 3 cos cos ( π + π) ⎝ 3π ⎠ cos x cos (3x − π)(3x + π) 3 3 3

lim sin(cos x ) cos x lim sin(cos x ) π = x→ π cos x 2 sin x − cos ecx 2

x→

lim

π x→ 2

lim cos 2 x sin x cos 2 x = 1 × x→ π 2 sin x − cos ecx sin 2 x − 1

sin θ ⎛ ⎞ = 1⎟ ⎜ u sin g lim θ → 0 θ ⎝ ⎠

lim =–

x→

π 2

(sin x) = – 1

Page # 4.

(c)

⎛ x+a⎞ ⎛x−a⎞ − 2 sin⎜ ⎟ sin⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ lim cos x − cos a lim x →a x →a x−a ⎛x−a⎞ 2×⎜ ⎟ ⎝ 2 ⎠ ⎛ x−a⎞ sin⎜ ⎟ ⎝ 2 ⎠ ⎛x+a⎞ ⎟ lim ⎛ x − a ⎞ = – sin a = – xlim sin ⎜ →a ⎝ 2 ⎠ x →a ⎜ ⎟ ⎝ 2 ⎠

(d)

lim

a sin x − x sin a

x →a

ax 2 − a 2 x

= xlim →a

a sin x − x sin x + x sin x − x sin a (a − x ) sin x + x(sin x − sin a) = xlim →a ax( x − a) ax( x − a)

⎛x+a⎞ 2 cos⎜ ⎟ (a − x ) sin x sin x − sin a sin a ⎝ 2 ⎠ lim lim = xlim + = – + →a x →a x →a ax( x − a) a( x − a) 2a a2

=

sin a a2

+

( x − a) ⎤ ⎡ ⎢ sin 2 ⎥ ⎢ ⎥ ⎢ ( x − a) ⎥ 2 ⎣⎢ ⎦⎥

cos a a

Example : 8 x ⎛ ⎞ ⎜ u sin g lim a − 1 = log a ⎟ ⎜ ⎟ Evaluate the following limits : x →0 x ⎝ ⎠

−e x−a

(a)

x lim 2 − 2 x →1 x −1

(b)

x →a

(c)

x x x lim 6 − 2 − 3 + 1 x →0 sin 2 x

(d)

x x lim 3 − 5 x →0 x

lim e

x

a

Solution (a)

(b)

x x −1 lim 2 − 2 = 2 lim 2 − 1 = 2 log 2 x →1 x →1 x −1 x −1

lim e

x →a

= e

a

= e

a

−e x−a

x

a

lim e

x →a

(1) xlim →a

e

= xlim →a

a⎛

x− a

− 1 lim x − a x →a

⎜ e x − a − 1⎞⎟ ⎝ ⎠ x −a x− a x−a

( x − a) ( x − a)

(

x+ a

)

=

e

a

2 a

Page # 5.

(c)

x x x lim 6 − 2 − 3 + 1 x →0 sin 2 x

x2 (2 x − 1)3 x − 1) lim = x →0 sin2 x x2 2

2 x − 1 lim 3 x − 1 lim ⎛ x ⎞ ⎜ ⎟ = loge2 loge2 = xlim →0 x →0 x →0 ⎝ sin x ⎠ x x

(d)

⎡ 3 x − 1 5 x − 1⎤ x x 3 lim 3 − 5 = lim ⎢ − ⎥ = log 3 – log 5 = log x →0 x →0 ⎢ x x 5 x ⎣ ⎦⎥

Example : 9 1 ⎡ ⎤ ⎢u sin g lim (1 + x ) x = e⎥ Evaluate the following limits : x →0 ⎢⎣ ⎥⎦

(a)

lim

1

x →0

lim xcotπx

(b)

(1 − 2x ) x

x →1

Solution

(a)

(b)

1 lim x →0 (1 − 2x ) x

lim xcotπx

x →1

= e

−1 ⎞ ⎛ ⎜ (1 − 2x ) 2 x ⎟ lim = x →0 ⎜ ⎟ ⎝ ⎠

1 ⎤ ⎡ x −1 ⎥ ⎢ lim 1 x 1 ) + − = x →1 ⎢⎣ ⎥⎦

lim ( x −1) cot πx

x →1

lim

= e

x →1

1− x tan( π − πx )

−2

( x −1) cot πx

lim

= e

x →1

1 π − πx π tan( π − πx )

=

1 π e

tan θ ⎛ ⎞ = 1⎟ ⎜ u sin g lim θ → 0 θ ⎝ ⎠

Example : 10 Show that the limit of : ⎧2 x − 1 ; x ≤ 1 f(x) = ⎨ ; x >1 ⎩ x

at x = 1 exists

Solution f(x) = xlim (2x – 1) = 2 – 1 = 1 Left hand limit = xlim →1− →1− (we use f(x) = 2x – 1

Q while calculating limit at x = 1, we approach x = 1 from LHS i.e. x < 1)

Right hand limit = xlim f(x) = xlim (x) = 1 →1+ →1+ ⇒

L.H.L. = R.H.L. = 1. Hence limit exists

Page # 6.

Example : 11 Find whether the following limits exist or not : (a)

lim sin 1 x

x →0

(b)

lim x sin 1 x

x →0

Solution (a)

1 →∞. x As the angle θ approaches ∞ , sin θ oscillates by taking values between – 1 and + 1.

As x → 0,

1 Hence xlim sin is not a well defined finite number. →0 x ⇒ limit does not exist

(b)

lim x sin 1 = lim x lim sin 1 x →0 x →0 x x = 0 × (some quantity between – 1 and + 1) = 0 x →0

1 1 It can be easily seen that xlim x sin = xlim x sin =0 →0 + →0 − x x Hence the limit exists and is equal to zero (0)

Example : 12 Comment on the following limits : (a)

lim [x – 3]

x →1

(b)

lim | x | x

x →0

Solution (a)

Right Hand limit = xlim [x – 3] →1+ = hlim [1 + h – 3] = hlim [h – 2] →0 →0 = – 2 (because h – 2 is between – 1 and – 2) Left hand limit = xlim [x – 3] →1− = hlim [1 – h – 3] = hlim [–2 – h] →0 →0 = – 3 (because – h – 2 is between – 2 and – 3) Hence R.H.L. ≠ L.H.L. ⇒ limit does not exist.

(b)

|x| −x Left hand limit = xlim = xlim =–1 →0 − →0 − x x |x| x Right hand limit = xlim = xlim =+1 →0 + →0 + x x Hence R.H.L. ≠ L.H.L. ⇒ limit does not exist

Example : 13 Find a and b so that the function :

π ⎧ ; 0≤x< ⎪ x + a 2 sin x 4 ⎪⎪ π π ; ≤x≤ f(x) = ⎨ 2x cot x + b 4 2 ⎪ ⎪a cos 2x − b sin x ; π < x ≤ π 2 ⎩⎪

Page # 7.

is continuous for x ∈ [0, π] Solution At x = π/4 Left hand limit = lim− f(x) = lim− (x + a 2 sin x) =

π +a 4

Right hand limit = lim+ f(x) = lim+ (2x cot x + b) =

π +b 2

π x→ 4

x→

π x→ 4

π 4

x→

π 4

⎛π⎞ ⎛π⎞ π π f ⎜ ⎟ = 2 ⎜ ⎟ cot +b= +b 4 4 ⎝ ⎠ ⎝ ⎠ 4 2

for continuity, these three must be equal π π +a= +b ⇒ 4 2 At x = π/2

⇒

a–b=

π 4

.............(i)

Left hand limit = lim− (2x cot x + b) = 0 + b = b x→

π 2

Right hand limit = lim+ (a cos 2x – b sin x) = – a – b x→

π 2

⎛π⎞ f ⎜ ⎟ =0+b ⎝2⎠

for continuity, b = – a – b ⇒ a + 2b = 0

............(ii)

Solving (i) and (ii) for a and b, we get : b = –

π π ,a= 6 12

Example : 14 A function f(x) satisfies the following property f(x + y) = f(x) f(y). Show that the function is continuous for all values of x if it is continuous at x = 1 Solution As the function is continuous at x = 1, we have

lim f(x) = lim− f(x) = f(1) x →1

x →1−

⇒

lim f(1 – h) = lim f(1 + h) = f(1) h→0

h→0

using f(x + y) = f(x) f(y), we get

lim f(1) f(–h) = lim f(1) f(h) = f(1) h→0

⇒

h→0

⇒

h→0

lim f(–h) = lim f(h) = 1 h→0

..............(i)

Now consider some arbitrary point x = a Left hand limit = hlim f(a – h) = hlim f(a) f(–h) →0 →0 = f(a) hlim f(–h) = f(a) →0

............. using (i)

Right hand limit = hlim f(a + h) = hlim f(a) f(h) →0 →0 = f(a) hlim f(h) = f(a) →0

............. using (i)

Hence at any arbitrary point (x = a) L.H.L. = R.H.L. = f(a) ⇒ function is continuous for all values of x.

Page # 8.

Example : 15 ⎧1 + x ; 0 ≤ x ≤ 2 f(x) = ⎨ ⎩3 − x ; 2 < x ≤ 3

Determine the form of g(x) = f(f(x)) and hence find the point of discontinuity of g, if any Solution

⎧ f (1 + x ) ; 0 ≤ x ≤ 1 ⎧1 + x ; 0 ≤ x ≤ 2 ⎪ g(x) = f(f(x)) = ⎨ = ⎨ f (1 + x ) ; 1 < x ≤ 2 3 − x ; 2 < x ≤ 3 ⎩ ⎪f ( 3 − x ) ; 2 < x ≤ 3 ⎩ Now

x ∈ [0, 1] x ∈ (0, 2] x ∈ (2, 3]

⇒ ⇒ ⇒

(1 + x) ∈ [1, 2] (1 + x) ∈ (2, 3] (3 – x) ∈ [0, 1)

Hence

⎧ f (1 + x ) for 0 ≤ x ≤ 1 ⇒ 1 ≤ x + 1 ≤ 2 ⎪ g(x) = ⎨ f (1 + x ) for 1 < x ≤ 2 ⇒ 2 < x + 1 ≤ 3 ⎪f (3 − x ) for 2 < x ≤ 3 ⇒ 0 ≤ 3 − x < 1 ⎩ Now if (1 + x) ∈ [1, 2], then f(1 + x) = 1 + (1 + x) = 2 + x [from the original definition of f(x)] Similarly if (1 + x) ∈ (2, 3), then f(1 + x) = 3 – (1 + x) = 2 – 2 .............(ii) If (3 – x) ∈ (0, 1), then f(3 – x) = 1 + (3 + x) = 4 – x .............(iii)

...........(i)

⎧2 + x ; 0 ≤ x ≤ 1 ⎪ Using (i), (ii) and (iii), we get g(x) = ⎨2 − x ; 1 < x ≤ 2 ⎪4 − x ; 2 < x ≤ 3 ⎩ Now we will check the continuity of g(x) at x = 1, 2 At x = 1 L.H.L. = xlim g(x) = xlim (2 + x) = 3 →1− →1− R.H.L. = xlim g(x) = xlim (2 – x) = 1 →1+ →1+ As L.H.L., g(x) is discontinuous at x = 1 At x = 2 L.H.L. = xlim g(x) = xlim (2 – x) = 0 →2 + →2 − R.H.L. = xlim g(x) = xlim (4 – x) = 2 →2 + →2 + As L.H.L. ≠ R.H.L., g(x) is discontinuous at x = 2 Example : 16 ⎧ e1/ x − 1 ⎪ ; x≠0 1/ x Discuss the continuity of f(x) = ⎨ e + 1 at the point x = 0 ⎪ 0 ; x=0 ⎩

Solution

LHL = xlim →0 −

1 ex

−1

1 ex

+1

t lim e − 1 = 0 − 1 = – 1 = t→ −∞ e t + 1 0 +1

Page # 9.

1

RHL = xlim →0 +

ex −1 1

ex +1

−t et − 1 lim 1 − e = tlim = →∞ e t + 1 t →∞ 1 + e − t

1− 0 =1 1+ 0

⇒

R.H.L. =

⇒

L.H.L. ≠ R.H.L. ⇒

f(x) is discontinuous at x = 0

Example : 17 Discuss the continuity of the function g(x) = [x] + [–x] at integral values of x. Solution Let us simplify the definition of the function (i) If x is an integer : [x] = x and [–x] = – x ⇒ g(x) = x – x = 0 (ii) If x is not integer : let x = n + f where n is an integer and f ∈ (0, 1) ⇒ [x] = [n + f] = n and [–x] = [–n – f] = [(–n – 1) + (1 – f)] = – n – 1 (because 0 < f < 1 ⇒ 0 , (1 – f) < 1) Hence g(x) = [x] + [–x] = n + (–n – 1) = – 1 So we get :

if x is an int eger ⎧0 , g(x) = ⎨ ⎩− 1 , if x is not an int eger

Let us discuss the continuity of g(x) at a point x = a where a ∈ Ι g(x) = – 1 L.H.L. = xlim →a − as x → a– , x is not an integer

Q

R.H.L. = xlim g(x) = – 1 →a + as x → a+ , x is not an integer but g(a) = 0 because a is an integer Hence g(x) has a removable discontinuity at integral values of x. Example : 18 Which of the following functions are even/odd? (a)

f(x) =

x

(b)

⎛ 1+ x ⎞ ⎟ f(x) = x log ⎜ ⎝ 1− x ⎠

(c)

f(x) = |x|

(d)

2 ⎛ ⎞ f(x) = log ⎜ x + x + 1 ⎟ ⎝ ⎠

ax − 1 a +1

Solution a −x − 1

= – f(x) ⇒ f(x) is odd

f(–x) =

(b)

⎛ 1− x ⎞ ⎛ 1− x ⎞ ⎟ = x log ⎜ ⎟ f(–x) = – x log ⎜ ⎝ 1+ x ⎠ ⎝ 1+ x ⎠

(c)

⇒ f(x) is even f(–x) = |–x| = |x| = f(x) ⇒ f(x) is even

a−x + 1

=

1− ax

(a)

1+ ax

−1

⎛ 1+ x ⎞ ⎟ = f(x) = x log ⎜ ⎝ 1− x ⎠

Page # 10.

(d)

⎛ 1+ x2 − x2 ⎜ ⎛ − x + 1+ x2 ⎞ ⎟ = log ⎜ f(–x) = log ⎜ 2 ⎝ ⎠ ⎝ x + 1+ x

⇒

⎞ ⎟ ⎟ = – log ⎠

⎛ x + x 2 + 1⎞ ⎜ ⎟ = – f(x) ⎝ ⎠

f(x) is odd.

Example : 19 Which of the following functions are periodic ? Give reasons (a)

f(x) = x + sin x

(b)

cos

x 2

(c) f(x) = x – [x] (d) cos x Solution If a function f(x) is periodic, then there should exist some positive value of constant a for which f(x + a) = f(x) is an identity (i.e. true for all x) The smallest value of a satisfying the above condition is known as the period of the function (a) Assume that f(x + a) = f(x) ⇒ x + a + sin (x + a) = x + sin x ⇒ sin x – sin (x + a) = a

(b)

⇒

a⎞ ⎛ a 2 cos ⎜ x + ⎟ sin =–a 2⎠ ⎝ 2

⇒

a⎞ ⎛ a 2 cos ⎜ x + ⎟ sin = – a 2⎠ ⎝ 2

This cannot be true for all values of x. Hence f(x) is non-periodic Assume that f(x + a) = f(x) ⇒

(c)

(d)

cos

x + a = cos

⇒

x + a = 2np ±

⇒

x+a ±

⇒

2x + a ± 2

⇒

2x ± 2

x x

x = 2nπ 2 2 x 2 + ax = 4n π

2 2 x 2 + ax = 4n π – a

As this equation cannot be an identity, 3 f(x) is non-periodic Assume that f(x + a) = f(x) ⇒ x + a – [x + a] = x – [x] ⇒ [x + a] – [x] = a This equation is true for all values of x if a is an integer hence f(x) is periodic Period = smallest positive value of a = 1 Let f(x + a) = f(x) ⇒ cos2 (x + a) = cos2x ⇒ cos2 (x + a) – cos2x = 0 ⇒ sin (2x + a) sin (a) = 0 this equation is true for all values of x if a is an integral multiple of π Hence f(x) is periodic. Period = smallest positive value of a = π

Example : 20 n

Find the natural number a for which

∑ f (a + k ) = 16(2

n

– 1) where the function f satisfies the relation

k =1

f(x + y) = f(x) f(y) for all natural numbers x, y and further f(1) = 2. Solution Since the function f satisfies the relation f(x + y) = f(x) f(y) It must be an exponential function.

Page # 11.

Let the base of this exponential function be a. Thus f(x) = ax It is given that f(1) = 2. So we can make ⇒ a=2 f(1) = a1 = 2 Hence, the function is f(x) = 2x ..........(i) [Alternatively, we have f(x) = f(x – 1 + 1) = f(x – 1) f(1) = f(x – 2 + 1) f(1) = f(x – 2) [f(1)]2 = .............. = [f(1)]x = 2x] Using equation (i), the given expression reduces to : n

∑2

a+k

k =1

= 16 (2n – 1)

n

⇒

∑2

a

k =1

. 2k = 16 (2n – 1)

n

∑2

k

⇒

2

⇒

2a (2 + 4 + 8 + 16 + ..............+ 2n) = 16 (2n – 1)

⇒

⎡ 2(2n − 1) ⎤ 2 ⎢ 2 − 1 ⎥ = 16 (2n – 1) ⎢⎣ ⎥⎦

⇒ ⇒

2a+1 = 16 a+1=4

a

k =1

= 16 (2n – 1)

a

⇒ ⇒

2a+1 = 24 a=3

Example : 21 Evaluate the following limits :

(i)

x 3− x − 12 lim 3 + 3 3− x x/2 x →2 3 −3

(ii)

lim

x→π / 3

tan3 x − 3 tan x π⎞ ⎛ cos⎜ x + ⎟ 6⎠ ⎝

(iii)

lim

x → −∞

1 + x2 x 1+ | x 3 |

x 4 sin

Solution (i)

3 x + 3 3 − x − 12 lim Let L = x →2 3 3− x − 3 x / 2

3x + ⇒

L = xlim →2

⇒

L = xlim →2

⇒

L = xlim →0

⇒

L = xlim →2

⇒

L=

27 − 12 3x

27 − 3x / 2 3x 3 2 x − 12.3 x + 27 (3 x / 2 )3 − 3 3 (3 x − 9)(3 x − 3) (3 x / 2 − 3)(3 x + 9 + 3.3 x / 2 ) (3 x / 2 + 3)(3 x − 3) ( 3 x + 3. 3 x / 2 + 9 )

6 .6 4 36 = = 9 + 3 .3 + 9 3 27

Page # 12.

(ii)

lim Let L = L = x→ π/3

⇒

(iii)

tan3 x − 3 tan x π⎞ ⎛ cos⎜ x + ⎟ 6 ⎝ ⎠

and

x–

π =t 3

π⎞ π⎞ ⎛ ⎛ tan3 ⎜1 + ⎟ − 3 tan⎜ t + ⎟ 3 3 ⎝ ⎠ ⎝ ⎠ L = tlim π ⎛ ⎞ →0 cos⎜ t + ⎟ ⎝ 2⎠

⇒

⎡ π⎞ ⎤ ⎛ tan(3t + π) ⎢3 tan 2 ⎜1 + ⎟ − 1⎥ ⎝ 3⎠ ⎦ ⎣ L = tlim →0 − sin t

⇒

− tan( 3t ) lim ⎡3 tan 2 ⎛ t + π ⎞ − 1⎤ ⎜ ⎟ ⎥ L = tlim . t →0 ⎢ →0 − sin t ⎝ 3⎠ ⎦ ⎣

⇒

⎡ π⎞ ⎤ tan( 3t ) 1 2⎛ ⎢3 tan ⎜1 + ⎟ − 1⎥ L = 3 tlim × tlim × tlim →0 → 0 → 0 3⎠ ⎦ 3t sin t ⎝ ⎣

⇒

L = 3 × 1 × 1 × 8 = 24

Let

L = xlim →−∞

1 + x2 x 1 + | x3 |

x 4 sin

Divided Numerator and Denominator by x3 to get

L = xlim →−∞

⇒

1 sin 1 x +1 2 x sin + x 1/ x x x = 1 ( − x )3 1 3 +|x| + x3 x3 x3

L = xlim →−∞

(Q

for x < 0, |x3| = – x3)

→ (1)+ → (0) =–1 → (0) + ( −1)

Example : 22

⎧ π a /|sin x| ; <x<0 ⎪(1+ | sin x | 6 ⎪ b ; x=0 Let f(x) = ⎨ tan 2 x ⎪ π ⎪ e tan 3 x ; 0<x< 6 ⎩ Determine a and b such that f(x) is continuous at x = 0 Solution Left hand limit at x = 0 a ⎤ ⎡ lim ( ) + 1 | sin x | |sin x| ⎥ L.H.L. = xlim f(x) = ⎢ →0 − x →0 − ⎣ ⎦

⇒

L.H.L. = hlim f(0 – h) →0

⇒

a ⎤ ⎡ ( 1+ | sinh |)|sinh| ⎥ = ex L.H.L. = hlim ⎢ →0 ⎣ ⎦

1 ⎡ ⎤ ⎢u sin g : lim(1 + t ) t = e⎥ t →0 ⎢⎣ ⎥⎦

Page # 13.

Right hand limit x = 0 tan 2 x

tan 3 x R.H.L. = xlim f(x) = xlim →0 + →0 + e

⇒

R.H.L. = hlim f(0 + h) →0

⇒

tan 3h R.H.L. = hlim →0 e

⇒

3⎝ R.H.L. = hlim →0 e

tan 2h

2 ⎛ tan 2h 3h ⎞ . ⎜ ⎟ 2h tan 3h ⎠

2

= e3

for continuity L.H.L. = R.H.L. = f(0) 2

⇒

ex = e 3 = b

⇒

a=

2 , b = e2/3 3

Example : 23 πx ⎞ ⎛ ⎜ sin ⎟ Discuss the continuity of f(x) in [0, 2] where f(x) = nlim →∞ 2 ⎠ ⎝

2n

Solution ⎧0 ; | x | < 1 Since nlim x2n = ⎨ →∞ ⎩1 ; | x | = 1

∴

πx ⎞ ⎛ ⎜ sin ⎟ f(x) = nlim →∞ 2 ⎠ ⎝

⎧ ⎪0 ; ⎪ = ⎨ ⎪1 ; ⎪⎩

2n

πx <1 2 πx =1 sin 2 sin

Thus f(x) is continuous for all x, except for those values of x for which sin

πx =1 2

i.e. x is an odd integer ⇒ x = (2n + 1) where x ∈ Ι Check continuity at x = (2n + 1) : L.H.L. =

lim

x →2n +1

f(x) = 0

............(i)

and f(2n + 1) = 1 ............(ii) from (i) and (ii), we get : L.H.L. ≠ f(2n + 1), ⇒ f(x) is discontinuous at x = 2n + 1 (i.e. at odd integers) Hence f(x) is discontinuous at x = (2n + 1).

Page # 14.

Example : 24 ⎧ ⎪ 1 − cos 4 x ; x<0 2 ⎪ x ⎪ a ; x=0 Let f (x) = ⎨ ⎪ x ; x>0 ⎪ ⎩⎪ 16 + x − 4

Determine the value of a, if possible, so that the function is continuous at x = 0. Solution ⎧ ⎪ 1 − cos 4 x ; x<0 2 ⎪ x ⎪ a ; x=0 It is given that f(x) = ⎨ ⎪ x ; x>0 ⎪ ⎩⎪ 16 + x − 4

is continuous at x = 0. So we can take :

lim f(x) = f(0) = lim+ f(x) x →0

x →0 −

Left hand limit at x = 0 1− cos 4 x f(x) = tlim L.H.L. = tlim →0 − →0 − x2

Now,

L.H.L. = hlim f(0 – h) →0

⇒

2 1− cos 4h lim 2 sin 2h = 8 L.H.L. = hlim = →0 h→0 h2 h2

sin t ⎤ ⎡ = 1⎥ ⎢u sin g : tlim → 0 t ⎦ ⎣

Right hand limit at x = 0 R.H.L. = tlim f(x) = tlim →0 + →0 +

x 16 + x − 4

Now,

R.H.L. = hlim f(0 + h) →0

⇒

R.H.L. = hlim →0

h 16 + h − 4

Rationalising denominator to get : ⇒

R.H.L. = hlim →0

h ⎛ ⎞ ⎜ 16 + h + 4 ⎟ = 8 ⎠ h ⎝

For function f(x) to be continuous at x = 0, L.H.L. = R.H.L. = f(0) ⇒ 8=8=a ⇒ a=8 Example : 25 Ler {x} and [x] denote the fractional and integral part of a real number x respectively. Solve 4{x} = x + [x]. Solution We can write x as : x = integral part + fractional part ⇒ x = [x] + {x} The given equation is 4{x} = x + [x]

Page # 15.

⇒ 4{x} = [x] + {x} + [x] ⇒ 3{x} = 2[x] .............(i) ⇒ 3{x} is an even integer But we have 0 ≤ {x} < 1 ⇒ 0 ≤ 3 {x} < 3 ⇒ 3{x} = 0, 2 [become 3{x} is even} ⇒

{x} = 0,

⇒ ⇒

2 3

3 2

and

[x] =

{x} = 0, 1

x=0+0

or

x=

2 +1 3

x=0

or

x=

5 3

........... (using i)

Example : 26 ; x ≤1 ⎧ [cos πx ] Discuss the continuity of f(x) in [0, 2] where f(x) = ⎨ ⎩| 2x − 3 | [ x − 2] ; x > 1

where [ ] : represents the greatest integer function. Solution First of all find critical points where f(x) may be discontinuous Consider x – [0, 1] : f(x) = [cos πx] [f(x)] is discontinuous where f(x) ∈ Ι ⇒ cos πx = Ι In [0, 1], cos πx is an integer at x = 0, x =

1 and x = 1 2

1 and x = 1 are critical points 2 Consider x – (1, 2] : f(x) = [x – 2] |2x – 3| In x ∈ (1, 2) [x – 2] = – 1 and for x = 2 ; [x – 2] = 0

⇒

x = 0, x =

⇒

Also |2x – 3| = 0 ⇒

x=

x=

............(i)

3 2

3 and x = 2 are critical points 2

............(i)

1 3 , 1, , 2 2 2 On dividing f(x) about the 5 critical points, we get

combining (i) and (ii), critical points are 0,

1 ⎧ ⎪ 0 ⎪ ⎪ ⎪ −1 ⎪ f(x) = ⎨ ⎪− 1(3 − 2x ) ⎪ ⎪ ⎪− 1(2 x − 3) ⎪ 0 ⎩

; ; ; ; ; ;

x=0

cos( π0 ) = 1 Q 1 0<x≤ 0 ≤ cos πx < 0 ⇒ [cos πx] = 0 Q 2 1 < x ≤ 1 Q − 1 ≤ cos πx < 0 ⇒ [cos πx ] = −1 2 3 1< x ≤ Q | 2x − 3 | = 3 − 2x and [ x − 2] = −1 2 3 < x < 2 Q | 2x − 3 | = 2x − 3 and [ x − 2] = −1 2 Q 2 [ x − 2] = 0

Page # 16.

Checking continuity at x = 0 : R.H.L. = xlim (0) = 0 →0 +

and

f(0) = 1

As, R.H.L. ≠ f(x) ⇒ f(x) is discontinuous at x = 0 Checking continuity at a + x = 1/2 L.H.L. = lim − f(x) = 0 x→

1 2

R.H.L. = lim f(x) = – 1 − x→

1 2

As L.H.L. ≠ R.H.L., 1 . 2 Checking continuity at x = 1 :

f(x) is discontinuous at x =

L.H.L. = xlim f(x) = – 1 →1− R.H.L. = xlim f(x) = – 1 = xlim (2x – 3) = – 1 →1+ →1+ and f(1) = – 1 As L.H.L = R.H.L. = f(1) f(x) is continuous at x = 1 Checking continuity at x = 3/2 : L.H.L. = lim − (2x – 3) = 0 x→

3 2

R.H.L. = lim + (3 – 2x) = 0 x→

As

and

⎛3⎞ f ⎜ ⎟ =0 ⎝2⎠

and

f(2) = 0

3 2

⎛3⎞ L.H.L. = R.H.L. = f ⎜ ⎟ , ⎝2⎠

f(x) is continuous at x = 3/2 Checking continuity at x = 2 : L.H.L. = xlim (3 – 2x) = – 1 →2 − As L.H.L. ≠ f(2), f(x) is discontinuous at x = 2 Example : 27 If f(x) =

sin 2 x + A sin x + B cos x x3

is continuous at x = 0, find the values of A and B. Also find f(0).

Solution As f(x) is continuous at x = 0 f(0) = xlim f(x) and both f(0) and xlim f(x) = are finite →a →a 2 sin 2x + A sin x + B cos x f(0) = xlim →0 x3 As denominator → 0 as x → 0, ∴ Numerator should also → 0 as x → 0. Which is possible only if (for f(0) to be finite) sin 2(0) + A sin (0) + B cos 0 = 0 ⇒ B=0

⇒

Page # 17.

∴

sin 2x + A sin x f(0) = xlim →0 x2

⇒

⎛ sin x ⎞ ⎛ 2 cos x + A ⎞ ⎛ 2 cos x + A ⎞ ⎜ ⎟ ⎜ ⎟ = lim ⎜ ⎟ f(0) = xlim 2 →0 ⎝ x ⎠ ⎝ x →0 ⎝ x x2 ⎠ ⎠

Again we can see that Denominator → 0 as x → 0 ∴ Numerator should also approach 0 as x → 0 ⇒ 2+A=0 ⇒ A=–2

⇒

(for f(0) to be finite)

⎞ ⎛ ⎛ 2 x ⎞ ⎜ − sin 2 x ⎟ ⎟ ⎜ − 4 sin 2⎟ ⎜ 2⎟ ⎛ 2 cos x − 2 ⎞ ⎜ lim ⎜ 2 ⎜ ⎟ = lim ⎜ 2 ⎟ =–1 f(0) = xlim = ⎟ 2 x x →0 ⎝ x →0 x ⎠ ⎜ ⎟ x →0 ⎜ ⎟ ⎠ ⎝ 4 ⎠ ⎝

So we get A = – 2, B = 0 and f(0) = – 1 Example : 28 Evaluate xlim →0

64 x − 32 x − 16 x + 4 x + 2 x − 1

( 3 + cos x − 2)sin x

Solution Let L = xlim →0

64 x − 32 x − 16 x + 4 x + 2 x − 1

( 3 + cos x − 2)sin x

On rationalising the denominated, we get L = xlim →0

2 6 x − 25 x − 2 4 x + 22 x + 2 x − 1 (cos x − 1) sin x

( 3 + cos x + 2)

On factorising the numerator, we get (2 x − 1)[2 5 x − 2 5 x (2 x − 1) + 1 L = xlim × xlim →0 →0 (cos x − 1) sin x

( 3 + cos x + 2)

⇒

( 2 x − 1)[(2 5 x − 2 3 x ) − ( 2 2 x − 1)] L = xlim ×4 →0 (cos x − 1) sin x

⇒

( 2 x − 1)( 2 2 x − 1)(2 3 x − 1) L = xlim ×4 →0 (cos x − 1) sin x

⇒

⎛ 2 x − 1⎞ ⎛ 22x − 1 ⎞ ⎜ ⎟ ⎟ lim ⎜ L = 4 xlim × 2 ×3 ⎟ →0 ⎜ x →0 ⎜ 2x ⎟ ⎝ x ⎠ ⎝ ⎠ ⎛ 23 x − 1 ⎞ ⎛ ⎞ x2 ⎛ x ⎞ ⎜ ⎟ ⎟ lim ⎜ lim lim ⎜ ⎟ × × 2 ⎜ ⎜ ⎟ ⎟ x →0 x →0 x →0 ⎝ sin x ⎠ ⎝ 3x ⎠ ⎝ − 2 sin ( x / 2) ⎠

⇒ Example : 29 (i)

L = 4 (ln 2) = 2 (ln 2) × 3 (ln 2) × (–2)

⇒

L = – 48 (ln 2)3

If f is an even function defined on the interval (–5, 5), then find the four real values of x satisfying ⎛ x +1⎞ ⎟. the equation f(x) = f ⎜ ⎝ x + 2⎠

(ii)

⎛ 1 + 5x 2 ⎜ Evaluate : xlim 2 →0 ⎜ ⎝ 1 + 3x

1

⎞ x2 ⎟ . ⎟ ⎠

Page # 18.

(iii)

π⎞ π⎞ ⎛ ⎛ ⎛5⎞ If f(x) = sin2x + sin2 ⎜ x + ⎟ = cos x cos ⎜ x + ⎟ and g ⎜ ⎟ = 1, then find g [f(x)]. 3⎠ 3⎠ ⎝ ⎝ ⎝4⎠

(iv)

Let f(x) = [x] sin

( π) where [ ] denotes the greater integer function. Find the domain of f(x) and [ x + 1]

the points of discontinuity of f(x) in the domain. Solution (i)

⎛ x +1⎞ ⎟ It is given that f(x) = f ⎜ ⎝ x + 2⎠

⇒

⎛ x +1⎞ ⎟ x= ⎜ ⎝ x + 2⎠

⇒

x=

⇒

x2 + x – 1 = 0

− 1± 5 2

...........(i)

As f(x) is even, f(x) = f(–x) ⎛ x +1⎞ ⎟ –x= ⎜ ⎝ x + 2⎠

⇒

x2 + 3x + 1 = 0 ⇒

x=

−3± 5 2

...........(ii)

One combining (i) and (ii), we get : x=

(ii)

− 1± 5 −3± 5 and x = . 2 2

⎛ 1 + 5x 2 ⎜ Let L = xlim 2 →0 ⎜ ⎝ 1 + 3x

1

⎞ x2 ⎟ ⎟ ⎠ 1

⇒

⎛ 1 + 5x 2 ⎞ x2 ⎜1 + ⎟ − 1 L = xlim 2 →0 ⎜ ⎟ ⎝ 1 + 3x ⎠

⇒

2 ⎛ ⎜1 + 2x L = xlim 2 →0 ⎜ ⎝ 1 + 3x

⇒

⎜ 1 + 3x ⎝ L = xlim →0 e

⎛ ⎜

(iii)

2

2

⎞ ⎟ ⎟ ⎠

1

⎞ x2 ⎟ ⎟ ⎠

2

=e

1 ⎡ ⎤ ⎢u sin g : lim(1 + t ) t = e⎥ t →0 ⎢⎣ ⎥⎦

π⎞ π⎞ ⎛ ⎛ It is given that f(x) = 1 – cos2x + sin2 ⎜ x + ⎟ + cos x cos ⎜ x + ⎟ 3⎠ 3⎠ ⎝ ⎝

⎡ π ⎞⎤ 2 2⎛ 1 ⎡2 cos x cos⎛⎜ x + π ⎞⎟⎤ ⎢ ⎥ = 1 – ⎢cos x − sin ⎜ x + 3 ⎟⎥ + 3 ⎠⎦ ⎝ ⎠⎦ ⎝ 2 ⎣ ⎣

π⎞ ⎛ π = 1 – cos ⎜ 2x + ⎟ cos + 3 3 ⎝ ⎠

π⎞ ⎛ π⎞ ⎛ π⎞ ⎛ cos⎜ ⎟ cos⎜ ⎟ cos⎜ 2x + ⎟ 3⎠ 5 ⎝3⎠ ⎝3⎠ ⎝ + =1+ = 2 2 2 4

Page # 19.

⇒

For all values of x, f(x) =

5 . (constant function) 4

⎛5⎞ Hence, g[f(x)] = g ⎜ ⎟ ⎝4⎠ ⎛5⎞ But g ⎜ ⎟ = 1 ⇒ ⎝4⎠

g [(f(x)] = 1

Hence, g[f(x)] = 1 for all values of x

(iv)

Let f(x) = [x] sin

( π) [ x + 1]

Domain of f(x) is x ∈ R excluding the point where [x + 1] = 0 (Q denominator cannot be zero) Find values of x which satisfy [x + 1] = 0 [x + 1] = 0 ⇒ 0≤x+1<1 ⇒ –1≤x<0 i.e. for all x ∈ [–1, 0), denominator is zero. So, domain is x ∈ R [–1, 0) ⇒ Domain is x ∈ (–∞, –1) ∪ [0, ∞) Point of Discontinuity As greatest integer function is discontinuous at integer points, f(x) is continuous for all non-integer points. Checking continuity at x = a (where a – 1) ⎛ ⎞ π ⎜⎜ ⎟⎟ L.H.L. = hlim [a – h] sin →0 ⎝ [a + 1 − h ] ⎠

⇒

⎛π⎞ L.H.L. = (a – 1) sin ⎜ ⎟ ⎝a⎠

............(i)

⎛ ⎞ π ⎜⎜ ⎟⎟ R.H.L. = hlim [a + h] sin →0 ⎝ [a + 1 + h] ⎠

⇒

⎛ π ⎞ ⎟ L.H.L. = a sin ⎜ ⎝ a + 1⎠

From (i) and (ii), L.H.L. ≠ R.H.L. ⇒ f(x) is discontinuous at x = a So, points of discontinuity are x ∈ Ι ∩ D. ⇒ x ∈ Ι – {–1}.

............(ii)

(i.e. at integer values of x) (i.e. integers lying in the set of domain)

Page # 20.