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

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

Solution:

We have,

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

=

=

=

=

=

=

=

= -2sina × (1/2)

= -sina

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

Solution:

We have,

lim_x→0[{x²-tan2x}/(tanx)]

Dividing numerator by 2x and denominator by x.

=

=

= 2(0 – 1)/1

= -2

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

Solution:

We have,

lim_x→0[{√2 – √(1 + cosx)}/x²]

On rationalizing numerator

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

= lim_x→0[(1-cosx))/x²{√2+√(1+cosx)}]

=

=

= 2 × (1/4) × [1/{√2 + √(1 + 1)}]

= (2/4) × (1/2√2)

= (1/4√2)

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

Solution:

We have,

lim_x→0[{xtanx}/(1 – cosx)]

On dividing the numerator and denominator by x²

=

=

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

= (4/2)

= 2

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

Solution:

We have,

lim_x→0[{x²+ 1 – cosx}/(xsinx)]

= lim_x→0[{x²+ 2sin²(x/2)}/(xsinx)]

On dividing the numerator and denominator by x²

=

=

As we know that lim_x→0[sinx/x] = 1

=

= 3/2

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

Solution:

We have,

lim_x→0[sin2x{cos3x – cosx}/(x³)]

=

=

As we know that lim_x→0[sinx/x] = 1

= -2 × 2 × 2

= -8

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

Solution:

We have,

lim_x→0[{2sinx°-sin2x°}/(x³)]

= lim_x→0[{2sinx°-2sinx°cosx°}/(x³)]

= lim_x→0[2sinx°{1-cosx°}/(x³)]

= lim_x→0[2sinx°{2sin²(x°/2)}/(x³)]

=

=

= 4 × [π³/(180 × 360 × 360)]

= (π/180)³

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

Solution:

We have,

lim_x→0[{x³.cotx}/(1 – cosx)]

= lim_x→0[x³/{tanx(1 – cosx)}]

=

=

=

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

= 2

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

Solution:

We have,

lim_x→0[{x.tanx}/(1 – cos2x)]

= lim_x→0[{x.tanx}/(2sin²x)]

On dividing the numerator and denominator by x²

=

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

= (1/2)

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

Solution:

We have,

lim_x→0[{sin(3 + x) – sin(3 – x)}/x]

=

= 2Lim_x→0[cos3.sinx/x]

= 2cos × 3lim_x→0[sinx/x]

As we know that lim_x→0[sinx/x] = 1

= 2cos3

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

Solution:

We have,

lim_x→0[{cos2x – 1)}/(cosx – 1)]

= lim_x→0[(2sin²x)/{2sin²(x/2)}]

= lim_x→0[(sin²x)/{sin²(x/2)}]

=

As we know that lim_x→0[sinx/x] = 1

= (x²) × (4/x²)

= 4

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

Solution:

We have,

lim_x→0[{3sin²x – 2sinx²)}/(3x²)]

= lim_x→0[(3sin²x/3x²) – (2sinx²/3x²)]

As we know that lim_x→0[sinx/x] = 1

= 1 – 2/3

= (3 – 2)/3

= (1/3)

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

Solution:

We have,

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

On rationalizing numerator.

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

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

=

As we know that lim_x→0[sinx/x] = 1

= 2 × {1/(√1 + √1)}

= 2/2

= 1

Question 45. lim_x→0[(1 – cos4x)/x²]

Solution:

We have,

lim_x→0[(1 – cos4x)/x²]

= lim_x→0[2sin²2x/x²]

=

As we know that lim_x→0[sinx/x] = 1

= 2 × 4

= 8

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

Solution:

We have,

lim_x→0[(xcosx + sinx)/(x²+ tanx)]

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

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

=

As we know that lim_x→0[tanx/x] = 1

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

= 2

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

Solution:

We have,

lim_x→0[(1 – cos2x)/(3tan²x)]

= limx→0[2sin²x/3tan²x]

=

= (2/3)lim_x→0[cos²x]

= (2/3)

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

Solution:

We have,

lim_θ→0[(1 – cos4θ)/(1 – cos6θ)]

= lim_θ→0[2sin²2θ/2sin²3θ]

= lim_θ→0[sin²2θ/sin²3θ]

=

= [(4θ²)/(9θ²)]

= (4/9)

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

Solution:

We have,

lim_x→0[(ax + xcosx)/(bsinx)]

On dividing the numerator and denominator by x

=

As we know that lim_x→0[sinx/x] = 1

=(a + cos 0)/b × 1

= (a + 1)/b

Question 50. lim_θ→0[(sin4θ)/(tan3θ)]

Solution:

We have,

lim_θ→0[(sin4θ)/(tan3θ)]

=

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

= (4θ/3θ)

= (4/3)

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

Solution:

We have,

lim_x→0[{2sinx – sin2x}/(x³)]

= lim_x→0[{2sinx – 2sinxcosx}/(x³)]

= lim_x→0[2sinx{1 – cosx}/(x³)]

= lim_x→0[2sinx{2sin²(x/2)}/(x³)]

=

As we know that lim_x→0[sinx/x] = 1

= (4/4)

= 1

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

Solution:

We have,

lim_x→0[{1 – cos5x}/{1 – cos6x}]

=

=

As we know that lim_x→0[sinx/x] = 1

= 25/(4 × 9)

= (25/36)

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

Solution:

We have,

lim_x→0[(cosecx – cotx)/x]

= lim_x→0[(1/sinx – cosx/sinx)/x]

= lim_x→0[(1 – cosx)/x.sinx]

= lim_x→0[2sin²(x/2)/x.sinx]

=

As we know that lim_x→0[sinx/x] = 1

= 2/4

= 1/2

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

Solution:

We have,

lim_x→0[(sin3x + 7x)/(4x + sin2x)]

=

=

=

As we know that lim_x→0[sinx/x] = 1

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

= 10/6

= 5/3

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

Solution:

We have,

lim_x→0[(5x + 4sin3x)/(4sin2x + 7x)]

=

=

=

As we know that lim_x→0[sinx/x] = 1

= (5 + 4 × 3)/(4 × 2 + 7)

= (17/15)

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

Solution:

We have,

lim_x→0[(3sinx – sin3x)/x³]

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

= lim_x→0[(4sin³x)/x³]

= 4Lim_x→0[{(sinx)/x}³]

As we know that lim_x→0[sinx/x] = 1

= 4 × 1

= 4

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

Solution:

We have,

lim_x→0[(tan2x – sin2x)/x³]

= lim_x→0[(sin2x/cos2x-sin2x)/x³]

=

=

= lim_x→0[(2sin2x.sin²x)/(x³cos2x)]

=

As we know that lim_x→0[sinx/x] = 1

= 2 × 2/cos0

= 4

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

Solution:

We have,

lim_x→0[(sinax + bx)/(ax + sinbx)]

=

=

=

As we know that lim_x→0[sinx/x] = 1

= (1 × a + b)/(a + 1 × b)

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

= 1

Question 59. lim_x→0[cosecx-cotx]

Solution:

We have,

lim_x→0[cosecx – cotx]

= lim_x→0[1/sinx – cosx/sinx]

= lim_x→0[(1 – cosx)/sinx]

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

= lim_x→0[sin(x/2)/cos(x/2)]

= lim_x→0[tan(x/2)/ x/2] × x/2

As we know that lim_x→0[tanx/x] = 1

= 0

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

Solution:

We have,

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

=

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

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

=

=

As we know that lim_x→0[sinx/x] = 1

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

= (1/0)

= ∞

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

Solution:

We have,

lim_x→0[(cosax – cosbx)/(cosecx – 1)]

=

=

=

= [(a + b)(a – b)/c²] × (4/4)

= (a²– b²)/c²

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

Solution:

We have,

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

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

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

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

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

=

= 0 + 2asina + 0 + a²cosa

= 2a + a²cosa

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

Solution:

We have,

lim_x→0[kx.cosecx] = lim_x→0[x.coseckx]

lim_x→0[kx/sinx] = lim_x→0[x/sinkx]

klim_x→0[x/sinx] = lim_x→0[kx/sinkx](1/k)

k = (1/k)

k²= 1

k = ±1

Article Tags :

Class 11

Mathematics

RD Sharma Solutions

School Learning

RD Sharma Class-11

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 45. limx→0[(1 – cos4x)/x2]

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

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

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

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

Question 50. limθ→0[(sin4θ)/(tan3θ)]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 45. lim_x→0[(1 – cos4x)/x²]

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

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

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

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

Question 50. lim_θ→0[(sin4θ)/(tan3θ)]

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

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

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

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

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

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

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

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

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

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

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

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