Решения RD Sharma класса 11 — Глава 29 Ограничения — Упражнение 29.7 | Набор 2

Опубликовано: 8 Октября, 2022

Вопрос 32. lim _x→0 [{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

Решение:

We have,
lim_x→0[{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]
=
=
=
=
=
=
=
= -2sina × (1/2)
= -sina

Вопрос 33. lim _x→0 [{x ² – tan2x}/(tanx)]

Решение:

We have,
lim_x→0[{x²-tan2x}/(tanx)]
Dividing numerator by 2x and denominator by x.
=
=
= 2(0 – 1)/1
= -2

Вопрос 34. lim _x→0 [{√2 – √(1 + cosx)}/x ² ]

Решение:

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)

Вопрос 35. lim _x→0 [{xtanx}/(1 – cosx)]

Решение:

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

Вопрос 36. lim _x→0 [{x ² + 1 – cosx}/(xsinx)]

Решение:

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

Вопрос 37. lim _x→0 [sin2x{cos3x – cosx}/(x ³ )]

Решение:

We have,
lim_x→0[sin2x{cos3x – cosx}/(x³)]
=
=
As we know that lim_x→0[sinx/x] = 1
= -2 × 2 × 2
= -8

Вопрос 38. lim _x→0 [{2sinx° – sin2x°}/(x ³ )]

Решение:

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

Вопрос 39. lim _x→0 [{x ³ .cotx}/(1 – cosx)]

Решение:

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

Вопрос 40. lim _x→0 [{x.tanx}/(1 – cos2x)]

Решение:

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)

Вопрос 41. lim _x→0 [{sin(3 + x) – sin(3 – x)}/x]

Решение:

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

Вопрос 42. lim _x→0 [{cos2x – 1)}/(cosx – 1)]

Решение:

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

Вопрос 43. lim _x→0 [{3sin ² x – 2sinx ² )}/(3x ² )]

Решение:

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)

Вопрос 44. lim _x→0 [{√(1 + sinx) – √(1 – sinx)}/x]

Решение:

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

Вопрос 45. lim _x→0 [(1 – cos4x)/x ² ]

Решение:

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

Вопрос 46. lim _x→0 [(xcosx + sinx)/(x ² + tanx)]

Решение:

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

Вопрос 47. lim _x→0 [(1 – cos2x)/(3tan ² x)]

Решение:

We have,
lim_x→0[(1 – cos2x)/(3tan²x)]
= limx→0[2sin²x/3tan²x]
=
= (2/3)lim_x→0[cos²x]
= (2/3)

Вопрос 48. lim _θ→0 [(1 – cos4θ)/(1 – cos6θ)]

Решение:

We have,
lim_θ→0[(1 – cos4θ)/(1 – cos6θ)]
= lim_θ→0[2sin²2θ/2sin²3θ]
= lim_θ→0[sin²2θ/sin²3θ]
=
= [(4θ²)/(9θ²)]
= (4/9)

Вопрос 49. lim _x→0 [(ax + xcosx)/(bsinx)]

Решение:

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

Вопрос 50. lim _θ→0 [(sin4θ)/(tan3θ)]

Решение:

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)

Вопрос 51. lim _x→0 [{2sinx – sin2x}/(x ³ )]

Решение:

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

Вопрос 52. lim _x→0 [{1 – cos5x}/{1 – cos6x}]

Решение:

We have,
lim_x→0[{1 – cos5x}/{1 – cos6x}]
=
=
As we know that lim_x→0[sinx/x] = 1
= 25/(4 × 9)
= (25/36)

Вопрос 53. lim _x→0 [(cosecx – cotx)/x]

Решение:

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

Вопрос 54. lim _x→0 [(sin3x + 7x)/(4x + sin2x)]

Решение:

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

Вопрос 55. lim _x→0 [(5x + 4sin3x)/(4sin2x + 7x)]

Решение:

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)

Вопрос 56. lim _x→0 [(3sinx – sin3x)/x ³ ]

Решение:

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

Вопрос 57. lim _x→0 [(tan2x – sin2x)/x ³ ]

Решение:

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

Вопрос 58. lim _x→0 [(sinax + bx)/(ax + sinbx)]

Решение:

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

Вопрос 59. lim _x→0 [cosecx-cotx]

Решение:

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

Вопрос 60. lim _x→0 [{sin(α + β)x + sin(α – β)x + sin2αx}/{cos ² βx – cos ² αx}]

Решение:

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)
= ∞

Вопрос 61. lim _x→0 [(cosax – cosbx)/(cosecx – 1)]

Решение:

We have,
lim_x→0[(cosax – cosbx)/(cosecx – 1)]
=
=
=
= [(a + b)(a – b)/c²] × (4/4)
= (a²– b²)/c²

Вопрос 62. lim _h→0 [{(a + h) ² sin(a + h) – a ² sina}/h]

Решение:

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

Вопрос 63. Если lim _x→0 [kx.cosecx] = lim _x→0 [x.cosecx], найти K.

Решение:

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

Решения RD Sharma класса 11 — Глава 29 Ограничения — Упражнение 29.7 | Набор 2

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

Вопрос 33. lim x→0 [{x 2 – tan2x}/(tanx)]

Вопрос 34. lim x→0 [{√2 – √(1 + cosx)}/x 2 ]

Вопрос 35. lim x→0 [{xtanx}/(1 – cosx)]

Вопрос 36. lim x→0 [{x 2 + 1 – cosx}/(xsinx)]

Вопрос 37. lim x→0 [sin2x{cos3x – cosx}/(x 3 )]

Вопрос 38. lim x→0 [{2sinx° – sin2x°}/(x 3 )]

Вопрос 39. lim x→0 [{x 3 .cotx}/(1 – cosx)]

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

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

Вопрос 42. lim x→0 [{cos2x – 1)}/(cosx – 1)]

Вопрос 43. lim x→0 [{3sin 2 x – 2sinx 2 )}/(3x 2 )]

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

Вопрос 45. lim x→0 [(1 – cos4x)/x 2 ]

Вопрос 46. lim x→0 [(xcosx + sinx)/(x 2 + tanx)]

Вопрос 47. lim x→0 [(1 – cos2x)/(3tan 2 x)]

Вопрос 48. lim θ→0 [(1 – cos4θ)/(1 – cos6θ)]

Вопрос 49. lim x→0 [(ax + xcosx)/(bsinx)]

Вопрос 50. lim θ→0 [(sin4θ)/(tan3θ)]

Вопрос 51. lim x→0 [{2sinx – sin2x}/(x 3 )]

Вопрос 52. lim x→0 [{1 – cos5x}/{1 – cos6x}]

Вопрос 53. lim x→0 [(cosecx – cotx)/x]

Вопрос 54. lim x→0 [(sin3x + 7x)/(4x + sin2x)]

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

Вопрос 56. lim x→0 [(3sinx – sin3x)/x 3 ]

Вопрос 57. lim x→0 [(tan2x – sin2x)/x 3 ]

Вопрос 58. lim x→0 [(sinax + bx)/(ax + sinbx)]

Вопрос 60. lim x→0 [{sin(α + β)x + sin(α – β)x + sin2αx}/{cos 2 βx – cos 2 αx}]

Вопрос 61. lim x→0 [(cosax – cosbx)/(cosecx – 1)]

Вопрос 62. lim h→0 [{(a + h) 2 sin(a + h) – a 2 sina}/h]

Вопрос 63. Если lim x→0 [kx.cosecx] = lim x→0 [x.cosecx], найти K.

РЕКОМЕНДУЕМЫЕ СТАТЬИ

Вопрос 32. lim _x→0 [{sin(a + x) + sin(a – x) – 2sina}/(xsinx)]

Вопрос 33. lim _x→0 [{x ² – tan2x}/(tanx)]

Вопрос 34. lim _x→0 [{√2 – √(1 + cosx)}/x ² ]

Вопрос 35. lim _x→0 [{xtanx}/(1 – cosx)]

Вопрос 36. lim _x→0 [{x ² + 1 – cosx}/(xsinx)]

Вопрос 37. lim _x→0 [sin2x{cos3x – cosx}/(x ³ )]

Вопрос 38. lim _x→0 [{2sinx° – sin2x°}/(x ³ )]

Вопрос 39. lim _x→0 [{x ³ .cotx}/(1 – cosx)]

Вопрос 40. lim _x→0 [{x.tanx}/(1 – cos2x)]

Вопрос 41. lim _x→0 [{sin(3 + x) – sin(3 – x)}/x]

Вопрос 42. lim _x→0 [{cos2x – 1)}/(cosx – 1)]

Вопрос 43. lim _x→0 [{3sin ² x – 2sinx ² )}/(3x ² )]

Вопрос 44. lim _x→0 [{√(1 + sinx) – √(1 – sinx)}/x]

Вопрос 45. lim _x→0 [(1 – cos4x)/x ² ]

Вопрос 46. lim _x→0 [(xcosx + sinx)/(x ² + tanx)]

Вопрос 47. lim _x→0 [(1 – cos2x)/(3tan ² x)]

Вопрос 48. lim _θ→0 [(1 – cos4θ)/(1 – cos6θ)]

Вопрос 49. lim _x→0 [(ax + xcosx)/(bsinx)]

Вопрос 50. lim _θ→0 [(sin4θ)/(tan3θ)]

Вопрос 51. lim _x→0 [{2sinx – sin2x}/(x ³ )]

Вопрос 52. lim _x→0 [{1 – cos5x}/{1 – cos6x}]

Вопрос 53. lim _x→0 [(cosecx – cotx)/x]

Вопрос 54. lim _x→0 [(sin3x + 7x)/(4x + sin2x)]

Вопрос 55. lim _x→0 [(5x + 4sin3x)/(4sin2x + 7x)]

Вопрос 56. lim _x→0 [(3sinx – sin3x)/x ³ ]

Вопрос 57. lim _x→0 [(tan2x – sin2x)/x ³ ]

Вопрос 58. lim _x→0 [(sinax + bx)/(ax + sinbx)]

Вопрос 60. lim _x→0 [{sin(α + β)x + sin(α – β)x + sin2αx}/{cos ² βx – cos ² αx}]

Вопрос 61. lim _x→0 [(cosax – cosbx)/(cosecx – 1)]

Вопрос 62. lim _h→0 [{(a + h) ² sin(a + h) – a ² sina}/h]

Вопрос 63. Если lim _x→0 [kx.cosecx] = lim _x→0 [x.cosecx], найти K.