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

  • Last Updated : 30 Apr, 2021

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

Solution:

We have,

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



\lim_{x\to0}[\frac{2sin(\frac{a+x+a-x}{2})sin(\frac{a+x-a+x}{2})}{xsinx}]

=\lim_{x\to0}[\frac{2sina(cosx-1)}{xsinx}]

=\lim_{x\to0}[\frac{-2sina(1-cosx)}{xsinx}]

-2sina\lim_{x\to0}[\frac{2sin^2\frac{x}{2}}{x.2sin\frac{x}{2}cos\frac{x}{2}}]

-2sina\lim_{x\to0}[\frac{sin\frac{x}{2}}{x.cos\frac{x}{2}}]

-2sina\lim_{x\to0}[\frac{tan\frac{x}{2}}{x}]

-2sina\lim_{x\to0}[\frac{tan\frac{x}{2}}{\frac{x}{2}×2}]

= -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.

=\lim_{x\to0}[\frac{(\frac{x^2}{2x}-\frac{tan2x}{2x})×2x}{\frac{tanx}{x}×x}]

=\lim_{x\to0}[\frac{2(\frac{x}{2}-\frac{tan2x}{2x})}{\frac{tanx}{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)}]

\lim_{x\to0}[\frac{sin^2(\frac{x}{2})}{x^2.(\sqrt{2}+\sqrt{1+cosx}}]

\lim_{x\to0}[\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2×4}×\frac{1}{(\sqrt{2}+\sqrt{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

=\lim_{x\to0}[\frac{\frac{tanx}{x}}{\frac{2sin^2(\frac{x}{2})}{x^2}}]

=\lim_{x\to0}[\frac{\frac{tanx}{x}}{\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2×4}}]

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

\lim_{x\to0}[\frac{1+\frac{2sin^2\frac{x}{2}}{x^2}}{{\frac{sinx}{x}}}]

\lim_{x\to0}[\frac{1+2(\frac{sin\frac{x}{2}}{\frac{x}{2}×2})^2}{{\frac{sinx}{x}}}]

As we know that limx→0[sinx/x] = 1 

\frac{1+\frac{2}{4}}{1}

= 3/2

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

Solution:



We have,

limx→0[sin2x{cos3x – cosx}/(x3)]

\lim_{x\to0}\frac{(sin2x)×[-2sin(\frac{3x+x}{2})sin(\frac{3x-x}{2})]}{x^3}

-2\lim_{x\to0}(\frac{sin2x}{2x})×(\frac{sin2x}{2x})×(\frac{sinx}{x})×2×2

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\lim_{x\to0}\frac{sin\frac{πx}{180}×sin^2(\frac{πx}{360})}{x.x^2}

4\lim_{x\to0}\frac{sin\frac{πx}{180}}{\frac{πx}{180}}×(\frac{sin(\frac{πx}{360})}{\frac{πx}{360}})^2×\frac{π}{180}×(\frac{π}{360})^2

= 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)}]

\lim_{x\to0}[\frac{x}{tanx}×\frac{x^2}{2sin^2\frac{x}{2}}]

\lim_{x\to0}[\frac{x}{tanx}×\frac{\frac{x^2}{4}×4}{2sin^2\frac{x}{2}}]

\lim_{x\to0}[\frac{x}{tanx}×(\frac{\frac{x}{2}}{sin\frac{x}{2}})^2×2]

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

\lim_{x\to0}[\frac{\frac{tanx}{x}}{2\frac{sin^2x}{x^2}}]

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]

=\lim_{x\to0}[\frac{2cos(\frac{3+x+3-x}{2})sin(\frac{3+x-3+x}{2})}{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)}]

\lim_{x\to0}[\frac{sin^2x}{x^2}×x^2][\frac{1}{\frac{sin^2\frac{x}{2}}{(\frac{x}{2})^2}×(\frac{x}{2})^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)}]

2\lim_{x\to0}[\frac{sinx}{x}×\frac{1}{\sqrt{1+sinx}+\sqrt{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]

2\lim_{x\to0}[(\frac{sin2x}{2x})^2×4]

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)]

\lim_{x\to0}\frac{\lim_{x\to0}cosx+\lim_{x\to0}\frac{sinx}{x}}{\lim_{x\to0}x+\lim_{x\to0}\frac{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]

=\frac{2}{3}\lim_{x\to0}\frac{sin^2x}{\frac{sin^2x}{cos^2x}}

= (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]

=\lim_{θ\to0}[\frac{sin^22θ}{(2θ)^2}×(2θ)^2×\frac{1}{\frac{sin^23θ}{(3θ)^2}×(3θ)^2}]

= [(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

\lim_{x\to0}\frac{(a+cosx)}{\frac{bsinx}{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)]

=\lim_{θ\to0}[\frac{sin4θ}{(4θ)}×(4θ)×\frac{1}{\frac{tan3θ}{(3θ)}×(3θ)}]

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)]

4\lim_{x\to0}\frac{sinx}{x}×\frac{sin^2\frac{x}{2}}{(\frac{x}{2})^2×4}

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}]

\lim_{x\to0}[\frac{2sin^2(\frac{5x}{x})}{2sin^23x}]

=\lim_{x\to0}[\frac{sin^2(\frac{5x}{2})}{(\frac{5x}{2})^2}×\frac{25x^2}{4}×\frac{1}{\frac{sin^23x}{(3x)^2}×9x^2}]

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]

2\lim_{x\to0}[\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}×(\frac{x^2}{4})×\frac{1}{\frac{xsinx}{x^2}×x^2}]

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)]

\lim_{x\to0}[\frac{\frac{sin3x}{3x}×3x+7x}{\frac{sin2x}{2x}×2x+4x}]

\lim_{x\to0}[\frac{x(\frac{sin3x}{3x}×3+7)}{x(\frac{sin2x}{2x}×2+4)}]



\lim_{x\to0}[\frac{(\frac{sin3x}{3x}×3+7)}{(\frac{sin2x}{2x}×2+4)}]

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)]

=\lim_{x\to0}[\frac{5x+4\frac{sin3x}{3x}×3x}{4\frac{sin2x}{2x}×2x+7x}]

=\lim_{x\to0}[\frac{x(5+4\frac{sin3x}{3x}×3)}{x(4\frac{sin2x}{2x}×2+7)}]

=\lim_{x\to0}[\frac{(5+4\frac{sin3x}{3x}×3)}{(4\frac{sin2x}{2x}×2+7)}]

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]

\lim_{x\to0}[\frac{sin2x(\frac{1}{cos2x}-1)}{x^3}]

\lim_{x\to0}[\frac{sin2x(1-cos2x)}{cos2x.x^3}]

= limx→0[(2sin2x.sin2x)/(x3cos2x)]



2\lim_{x\to0}[\frac{sin2x}{2x}×2×(\frac{sinx}{x})^2×\frac{1}{cos2x}]

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)]

\lim_{x\to0}[\frac{\frac{sinax}{ax}×ax+bx}{ax+\frac{sinbx}{bx}×bx}]

\lim_{x\to0}[\frac{x(\frac{sinax}{ax}×x+b)}{x(a+\frac{sinbx}{bx}×b)}]

\lim_{x\to0}[\frac{(\frac{sinax}{ax}×x+b)}{(a+\frac{sinbx}{bx}×b)}]

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}]

=\lim_{x\to0}[\frac{2sin(\frac{αx+βx+αx-βx}{2})cos(\frac{αx+βx-αx+βx}{2})+2sinαx.cosαx}{(1-sin^2βx)-(1-sin^2αx)}]

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

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

2\lim_{x\to0}[\frac{\frac{sinαx}{αx}×αx×(cosβx+cosαx)}{\frac{sin^2αx}{α^2x^2}×α^2x^2-\frac{sin^2βx}{β^2x^2}×β^2x^2}]

\frac{\lim_{x\to0}\frac{sinαx}{αx}\times \lim_{x\to0}(cosβx+cosαx)}{\lim_{x\to0}\frac{sin^2αx}{α^2x^2}×α^2x-\lim_{x\to0}\frac{sin^2βx}{β^2}×β^2x^2} \times \frac{x}{x^2}

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)]



=\lim_{x\to0}[\frac{-2sin(\frac{ax+bx}{2})sin(\frac{ax-bx}{2})}{-2sin^2(\frac{cx}{2})}]

=\lim_{x\to0}[\frac{sin(\frac{ax+bx}{2})sin(\frac{ax-bx}{2})}{sin^2(\frac{cx}{2})}]

=\lim_{x\to0}[\frac{\frac{sin(\frac{ax+bx}{2})}{(\frac{ax+bx}{2})}×(\frac{ax+bx}{2})×\frac{sin(\frac{ax-bx}{2})}{(\frac{ax-bx}{2})}×\frac{ax-bx}{2}}{(\frac{sin\frac{cx}{2}}{\frac{cx}{2}})^2×(\frac{cx}{2})^2}]

= [(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]

=\lim_{h\to0}[\frac{-a^2sina*sin^2(\frac{h}{2})}{\frac{h}{2}}+2asina.cosh+hsina.cosh+(a+h)^2cosa.sinh]

= 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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