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)]

= \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. lim_x→0[{x² - tan2x}/(tanx)]

Solution:

We have,

lim_x→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. 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)}]

= \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. lim_x→0[{xtanx}/(1 - cosx)]

Solution:

We have,

lim_x→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 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

= \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 lim_x→0[sinx/x] = 1 

= \frac{1+\frac{2}{4}}{1}

= 3/2

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

Solution:

We have,

lim_x→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 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\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. lim_x→0[{x³.cotx}/(1 - cosx)]

Solution:

We have,

lim_x→0[{x³.cotx}/(1 - cosx)]

= lim_x→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 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

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

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]

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

= \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 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)}]

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

= 2\lim_{x\to0}[(\frac{sin2x}{2x})^2×4]

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)]

= \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 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]

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

= (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]

=\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. lim_x→0[(ax + xcosx)/(bsinx)]

Solution:

We have,

lim_x→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 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)]

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

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)]

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

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}]

= \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 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]

= 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 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)]

= \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 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)]

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

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

= lim_x→0[(2sin2x.sin²x)/(x³cos2x)]

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

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)]

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

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

= lim_x→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 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)]

=\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. 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]

=\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 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