11类RD Sharma解决方案–第29章限制–练习29.7 |套装2

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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)

问题35. lim _x→0 [{xtanx} /(1-cosx)]

解决方案：

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

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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

问题40. lim _x→0 [{x.tanx} /(1-cos2x)]

解决方案：

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)

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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)

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

解决方案：

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

问题45. lim _x→0 [(1-cos4x)/ x ² ]

解决方案：

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

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

解决方案：

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

问题47. lim _x→0 [(1-cos2x)/(3tan ² x)]

解决方案：

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)

问题48. _limθ→0 [(1-cos4θ)/(1-cos6θ)]

解决方案：

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)

问题49. lim _x→0 [(ax + xcosx)/(bsinx)]

解决方案：

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

问题50. _limθ→0 [(sin4θ)/(tan3θ)]

解决方案：

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)

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

解决方案：

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

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

解决方案：

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)

问题53. lim _x→0 [(cosecx-cotx)/ x]

解决方案：

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

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

解决方案：

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

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

解决方案：

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)

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

解决方案：

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

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

解决方案：

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

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

解决方案：

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

问题59. lim _x→0 [cosecx-cotx]

解决方案：

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

问题60. LIM _X→0 [{SIN(α+β)X + SIN(α – β)X +sin2αx} / ^{2 COSβX – COS ^2αX}]

解决方案：

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)

= ∞

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

解决方案：

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

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

解决方案：

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

问题63.如果lim _x→0 [kx.cosecx] = lim _x→0 [x.coseckx]，请找到K。

解决方案：

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