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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma solutions – Chapter 29 Limits – Exercise 29.7 | Set 2

### Question 32. limx→0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

Solution:

We have,

limx→0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

=

=

= -2sina × (1/2)

= -sina

### Question 33. limx→0[{x2 – tan2x}/(tanx)]

Solution:

We have,

limx→0[{x2-tan2x}/(tanx)]

Dividing numerator by 2x and denominator by x.

=

=

= 2(0 – 1)/1

= -2

### Question 34. limx→0[{√2 – √(1 + cosx)}/x2]

Solution:

We have,

limx→0[{√2 – √(1 + cosx)}/x2]

On rationalizing numerator

= limx→0[{2-(1+cosx)}/x2{√2+√(1+cosx)}]

= limx→0[(1-cosx))/x2{√2+√(1+cosx)}]

= 2 × (1/4) × [1/{√2 + √(1 + 1)}]

= (2/4) × (1/2√2)

= (1/4√2)

### Question 35. limx→0[{xtanx}/(1 – cosx)]

Solution:

We have,

limx→0[{xtanx}/(1 – cosx)]

On dividing the numerator and denominator by x2

=

=

As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1

= (4/2)

= 2

### Question 36. limx→0[{x2 + 1 – cosx}/(xsinx)]

Solution:

We have,

limx→0[{x2 + 1 – cosx}/(xsinx)]

= limx→0[{x2 + 2sin2(x/2)}/(xsinx)]

On dividing the numerator and denominator by x2

As we know that limx→0[sinx/x] = 1

= 3/2

### Question 37. limx→0[sin2x{cos3x – cosx}/(x3)]

Solution:

We have,

limx→0[sin2x{cos3x – cosx}/(x3)]

As we know that limx→0[sinx/x] = 1

= -2 × 2 × 2

= -8

### Question 38. limx→0[{2sinx° – sin2x°}/(x3)]

Solution:

We have,

limx→0[{2sinx°-sin2x°}/(x3)]

= limx→0[{2sinx°-2sinx°cosx°}/(x3)]

= limx→0[2sinx°{1-cosx°}/(x3)]

= limx→0[2sinx°{2sin2(x°/2)}/(x3)]

= 4 × [π3/(180 × 360 × 360)]

= (π/180)3

### Question 39. limx→0[{x3.cotx}/(1 – cosx)]

Solution:

We have,

limx→0[{x3.cotx}/(1 – cosx)]

= limx→0[x3/{tanx(1 – cosx)}]

As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1

= 2

### Question 40. limx→0[{x.tanx}/(1 – cos2x)]

Solution:

We have,

limx→0[{x.tanx}/(1 – cos2x)]

= limx→0[{x.tanx}/(2sin2x)]

On dividing the numerator and denominator by x2

As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1

= (1/2)

### Question 41. limx→0[{sin(3 + x) – sin(3 – x)}/x]

Solution:

We have,

limx→0[{sin(3 + x) – sin(3 – x)}/x]

=

= 2Limx→0[cos3.sinx/x]

= 2cos × 3limx→0[sinx/x]

As we know that limx→0[sinx/x] = 1

= 2cos3

### Question 42. limx→0[{cos2x – 1)}/(cosx – 1)]

Solution:

We have,

limx→0[{cos2x – 1)}/(cosx – 1)]

= limx→0[(2sin2x)/{2sin2(x/2)}]

= limx→0[(sin2x)/{sin2(x/2)}]

As we know that limx→0[sinx/x] = 1

= (x2) × (4/x2)

= 4

### Question 43. limx→0[{3sin2x – 2sinx2)}/(3x2)]

Solution:

We have,

limx→0[{3sin2x – 2sinx2)}/(3x2)]

= limx→0[(3sin2x/3x2) – (2sinx2/3x2)]

As we know that limx→0[sinx/x] = 1

= 1 – 2/3

= (3 – 2)/3

= (1/3)

### Question 44. limx→0[{√(1 + sinx) – √(1 – sinx)}/x]

Solution:

We have,

limx→0[{√(1 + sinx) – √(1 – sinx)}/x]

On rationalizing numerator.

= limx→0[{(1 + sinx) – (1 – sinx)}/x{√(1 + sinx) + √(1 – sinx)}]

= limx→0[2(sinx)/x{√(1 + sinx) + √(1 – sinx)}]

As we know that limx→0[sinx/x] = 1

= 2 × {1/(√1 + √1)}

= 2/2

= 1

### Question 45. limx→0[(1 – cos4x)/x2]

Solution:

We have,

limx→0[(1 – cos4x)/x2]

= limx→0[2sin22x/x2]

As we know that limx→0[sinx/x] = 1

= 2 × 4

= 8

### Question 46. limx→0[(xcosx + sinx)/(x2 + tanx)]

Solution:

We have,

limx→0[(xcosx + sinx)/(x2 + tanx)]

= limx→0[x(cosx+sinx/x)/x(x + tanx/x)]

= limx→0[(cosx + sinx/x)/(x + tanx/x)]

As we know that limx→0[tanx/x] = 1

= (1 + 1)/(1 + 0)

= 2

### Question 47. limx→0[(1 – cos2x)/(3tan2x)]

Solution:

We have,

limx→0[(1 – cos2x)/(3tan2x)]

= limx→0[2sin2x/3tan2x]

=

= (2/3)limx→0[cos2x]

= (2/3)

### Question 48. limθ→0[(1 – cos4θ)/(1 – cos6θ)]

Solution:

We have,

limθ→0[(1 – cos4θ)/(1 – cos6θ)]

= limθ→0[2sin22θ/2sin23θ]

= limθ→0[sin22θ/sin23θ]

=

= [(4θ2)/(9θ2)]

= (4/9)

### Question 49. limx→0[(ax + xcosx)/(bsinx)]

Solution:

We have,

limx→0[(ax + xcosx)/(bsinx)]

On dividing the numerator and denominator by x

As we know that limx→0[sinx/x] = 1

=(a + cos 0)/b × 1

= (a + 1)/b

### Question 50. limθ→0[(sin4θ)/(tan3θ)]

Solution:

We have,

limθ→0[(sin4θ)/(tan3θ)]

=

As we know that limx→0[sinx/x] = 1 and limx→0[tanx/x] = 1

= (4θ/3θ)

= (4/3)

### Question 51. limx→0[{2sinx – sin2x}/(x3)]

Solution:

We have,

limx→0[{2sinx – sin2x}/(x3)]

= limx→0[{2sinx – 2sinxcosx}/(x3)]

= limx→0[2sinx{1 – cosx}/(x3)]

= limx→0[2sinx{2sin2(x/2)}/(x3)]

As we know that limx→0[sinx/x] = 1

= (4/4)

= 1

### Question 52. limx→0[{1 – cos5x}/{1 – cos6x}]

Solution:

We have,

limx→0[{1 – cos5x}/{1 – cos6x}]

=

As we know that limx→0[sinx/x] = 1

= 25/(4 × 9)

= (25/36)

### Question 53. limx→0[(cosecx – cotx)/x]

Solution:

We have,

limx→0[(cosecx – cotx)/x]

= limx→0[(1/sinx – cosx/sinx)/x]

= limx→0[(1 – cosx)/x.sinx]

= limx→0[2sin2(x/2)/x.sinx]

As we know that limx→0[sinx/x] = 1

= 2/4

= 1/2

### Question 54. limx→0[(sin3x + 7x)/(4x + sin2x)]

Solution:

We have,

limx→0[(sin3x + 7x)/(4x + sin2x)]

As we know that limx→0[sinx/x] = 1

= (7 + 3)/(4 + 2)

= 10/6

= 5/3

### Question 55. limx→0[(5x + 4sin3x)/(4sin2x + 7x)]

Solution:

We have,

limx→0[(5x + 4sin3x)/(4sin2x + 7x)]

=

=

=

As we know that limx→0[sinx/x] = 1

= (5 + 4 × 3)/(4 × 2 + 7)

= (17/15)

### Question 56. limx→0[(3sinx – sin3x)/x3]

Solution:

We have,

limx→0[(3sinx – sin3x)/x3]

= limx→0[{3sinx – (3sinx – 4sin3x)/x3]

= limx→0[(4sin3x)/x3]

= 4Limx→0[{(sinx)/x}3]

As we know that limx→0[sinx/x] = 1

= 4 × 1

= 4

### Question 57. limx→0[(tan2x – sin2x)/x3]

Solution:

We have,

limx→0[(tan2x – sin2x)/x3]

= limx→0[(sin2x/cos2x-sin2x)/x3]

= limx→0[(2sin2x.sin2x)/(x3cos2x)]

As we know that limx→0[sinx/x] = 1

= 2 × 2/cos0

= 4

### Question 58. limx→0[(sinax + bx)/(ax + sinbx)]

Solution:

We have,

limx→0[(sinax + bx)/(ax + sinbx)]

As we know that limx→0[sinx/x] = 1

= (1 × a + b)/(a + 1 × b)

= (a + b)/(a + b)

= 1

Question 59. limx→0[cosecx-cotx]

Solution:

We have,

limx→0[cosecx – cotx]

= limx→0[1/sinx – cosx/sinx]

= limx→0[(1 – cosx)/sinx]

= limx→0[{2sin2(x/2)}/{2sin(x/2)cos(x/2)}]

= limx→0[sin(x/2)/cos(x/2)]

= limx→0[tan(x/2)/ x/2] × x/2

As we know that limx→0[tanx/x] = 1

= 0

### Question 60. limx→0[{sin(α + β)x + sin(α – β)x + sin2αx}/{cos2βx – cos2αx}]

Solution:

We have,

limx→0[{sin(α + β)x + sin(α – β)x + sin2αx}/{cos2βx – cos2αx}]

=

= limx→0[{2sinαx.cosβx + 2sinαx.cosαx}/(sin2αx – sin2βx)]

= limx→0[{2sinαx(cosβx + cosαx)}/(sin2αx – sin2βx)]

As we know that limx→0[sinx/x] = 1

= [{2 × α × 1 × (1 + 1)}/(α2 – β2)] × (1/0)

= (1/0)

= ∞

### Question 61. limx→0[(cosax – cosbx)/(cosecx – 1)]

Solution:

We have,

limx→0[(cosax – cosbx)/(cosecx – 1)]

=

=

=

= [(a + b)(a – b)/c2] × (4/4)

= (a2 – b2)/c2

### Question 62. limh→0[{(a + h)2sin(a + h) – a2sina}/h]

Solution:

We have,

limh→0[{(a + h)2sin(a + h) – a2sina}/h]

= limh→0[{(a+h)2(sina.cosh)+(a+h)2(cosa.sinh)-a2sina}/h]

= limh→0[{(a2+2ah+h2)(sina.cosh)-a2sina+(a+h)2(cosa.sinh)}/h]

= limh→0[{a2sina(cosh-1)+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]

= limh→0[{a2sina(-2sin2(h/2))+2ah.sina.cosh+h2sina.cosh+(a+h)2cosa.sinh}/h]

=

= 0 + 2asina + 0 + a2cosa

= 2a + a2cosa

### Question 63. If limx→0[kx.cosecx] = limx→0[x.coseckx], find K.

Solution:

We have,

limx→0[kx.cosecx] = limx→0[x.coseckx]

limx→0[kx/sinx] = limx→0[x/sinkx]

klimx→0[x/sinx] = limx→0[kx/sinkx](1/k)

k = (1/k)

k2 = 1

k = ±1

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