11类RD Sharma解决方案–第29章限制–练习29.7 |套装1

问题1. Lim _x→0 [sin3x / 5x]

解决方案：

We have,

Lim_x→0[sin3x/5x]

= (1/5)Lim_x→0[sin3x/3x] × 3

= (3/5)Lim_x→0[sin3x/3x]

As we know that, Lim_x→0[sinx/x] = 1

= (3/5)           

问题2. Lim _x→0 [sinx°/ x]

解决方案：

We have,

Lim_x→0[sinx°/x]

As we know that x° = [(πx)/180]

= Lim_x→0[sin{(πx)/180}/x]

= $\lim_{x\to0}\frac{sin\frac{πx}{180}}{\frac{πx}{180}}×\frac{π}{180}$

= (π/180) × 1

= (π/180)

问题3. Lim _x→0 [x ² / sinx ² ]

解决方案：

We have,

Lim_x→0[x²/sinx²]

=$Lim_{x→0}[\frac{1}{\frac{sinx^2}{x^2}}]$

= $[\frac{1}{ Lim_{x→0}\frac{sinx^2}{x^2}}]$

As we know that, Lim_x→0[sinx/x] = 1

= 1/1

= 1

问题4. Lim _x→0 [sinx.cosx / 3x]

解决方案：

We have,

Lim_x→0[(sinx.cosx)/3x]

= Lim_x→0[(sinx.cosx)/ × 3x]

= 1/3 Lim_x→0[(sinx)/x] × Lim_x→0[(cosx)]

As we know that  Lim_x→0[sinx/x] = 1, and Lim_x→0cos0 = 1

= (1/3) × 1 × 1      

= 1/3

问题5. Lim _x→0 [(3sinx – 4sin ³ x)/ x]

解决方案：

We have,

Lim_x→0[(3sinx – 4sin³x)/x]

As we know that 3sinx – 4sin³x =  sin3x

= Lim_x→0[(sin3x)/3x] × 3

= 3 × Lim_x→0[(sin3x)/3x]

As we know that, Lim_x→0[sinx/x] = 1

= 3 × 1

= 3

问题6. Lim _x→0 [tan8x / sin2x]

解决方案：

We have,

Lim_x→0[tan8x/sin2x]

= $\lim_{x\to0}[\frac{tan8x}{8x}×8x][\frac{1}{\frac{sin2x}{2x}×2x}]$

= $\lim_{x\to0}[\frac{tan8x}{8x}][\frac{1}{\frac{sin2x}{2x}}]×\frac{8x}{2x}$

As we know that Lim_x→0[sin2x/2x] = 1 and Lim_x→0[tan8x/8x] = 1

= 8/2

= 4         

问题7. Lim _x→0 [tan(mx)/ tan(nx)]

解决方案：

We have,

Lim_x→0[tan(mx)/tan(nx)]

= $\lim_{x\to0}[\frac{tanmx}{mx}×mx][\frac{1}{\frac{tannx}{nx}×nx}]$

= $\lim_{x\to0}[\frac{tanmx}{mx}][\frac{1}{\frac{tannx}{nx}}]×\frac{mx}{nx}$

As we know that Lim_x→0[tanx/x] = 1

= m × 1/n × 1

= m/n           

问题8. Lim _x→0 [sin5x / tan3x]

解决方案：

We have,

Lim_x→0[sin5x/tan3x]

=$\lim_{x\to0}[\frac{sin5x}{5x}×5x][\frac{1}{\frac{tan3x}{3x}×3x}]$

=$\lim_{x\to0}[\frac{sin5x}{5x}][\frac{1}{\frac{tan3x}{3x}}]×\frac{5x}{3x}$

As we know that Lim_x→0[sin5x/5x] = 1 and Lim_x→0[tan3x/3x] = 1

= 5/3 × 1

= 5/3

问题9. Lim _x→0 [sin(x ⁿ )/(x ⁿ )]

解决方案：

We have,

Lim_x→0[sin(xⁿ)/(xⁿ)]

It is in the form of 0/0

So, let xⁿ = y

x→0 than y→0

Lim_y→0[siny/y]

As we know that Lim_y→0[siny/y] = 1

= 1         

问题10. Lim _x→0 [(7xcosx – 3sinx)/(4x + tanx)]

解决方案：

We have,

Lim_x→0[(7xcosx – 3sinx)/(4x + tanx)]

= Lim_x→0[x(7cosx-3sinx/x)/x(4 + tanx/x)]

= Lim_x→0[(7cosx-3sinx/x)/(4 + tanx/x)]

= $\frac{7\lim_{x\to0}cosx-3\lim_{x\to0}\frac{sinx}{x}}{4+\lim_{x\to0}\frac{tanx}{x}}$

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

= (7 – 3)/(4 + 1)

= 4/5           

问题11. Lim _x→0 [{cos(ax)– cos(bx)} / {cos(cx)– cos(dx)}]

解决方案：

We have,

Lim_x→0[{cos(ax) – cos(bx)}/{cos(cx) – cos(dx)}]

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

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

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

=$\lim_{x\to0}[\frac{x^2(a+b)(a-b)}{x^2(c+d)(c-d)}]$

= (a² – b²)/(c² – d²)

问题12. Lim _x→0 [(tan ² 3x)/ x ² ]

解决方案：

We have,

Lim_x→0[(tan²3x)/x²]

= Lim_x→0[(tan3x)/x]²

= Lim_x→0[(tan3x)/3x]² × 9

As we know that Lim_x→0[tanx/x] = 1

= 1 × 9

= 9

问题13. Lim _x→0 [(1-cosmx)/ x ² ]

解决方案：

We have,

Lim_x→0[(1 – cosmx)/x²]

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

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

= $2\lim_{x\to0}[\frac{sin\frac{mx}{2}}{\frac{mx}{2}}]^2×(\frac{m}{2})^2$

As we know that Lim_x→0[sinx/x] = 1

= 2 × (m/2)²

= m²/2

问题14. Lim _x→0 [(3sin2x + 2x)/(3x + 2tan3x)]

解决方案：

We have,

Lim_x→0[(3sin2x + 2x)/(3x + 2tan3x)]

= Lim_x→0[2x(1 + 3sin2x/2x)/3 x (1 + 2tan3x/3x)]

= (2/3)Lim_x→0[(1 + 3sin2x/2x)/3 x (1 + 2tan3x/3x)]

=$\frac{2}{3}\lim_{x\to0}\frac{1+3\lim_{x\to0}\frac{sin2x}{2x}}{1+2\lim_{x\to0}\frac{tan3x}{3x}}$

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

= (2/3) × (4/3)

= 8/9

问题15. Lim _x→0 [(cos3x – cos7x)/ x ² ]

解决方案：

We have,

=$\lim_{x\to0}[\frac{-2sin(\frac{3x+7x}{2})sin(\frac{3x-7x}{2})}{x^2}]$

= Lim_x→0[-2sin5x.sin(-2x)/x²]

= Lim_x→0[2sin5x.sin2x/x²]

= $2\lim_{x\to0}[\frac{sin5x}{5x}×5]×\lim_{x\to0}[\frac{sin2x}{2x}×2]$

As we know that Lim_x→0[sinx/x] = 1

= 2 × 5 × 2

= 20

问题16. _Limθ→0 [sin3θ/tan2θ]

解决方案：

We have,

Lim_θ→0[sin3θ/tan2θ]

= $\lim_{θ\to0}[\frac{sin3θ}{3θ}×3θ][\frac{1}{\frac{tan2θ}{2θ}×2θ}]$

= $\lim_{x\to0}[\frac{sin3θ}{3θ}][\frac{1}{\frac{tan2θ}{2θ}}]×\frac{3θ}{2θ}$

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

= 3/2

问题17. Lim _x→0 [sinx ² (1-cosx ² )/ x ⁶ ]

解决方案：

We have,

Lim_x→0[sinx²(1 – cosx²)/x⁶]

= Lim_x→0[sinx²{2sin(x²/2}/x⁶]

= $2\lim_{x\to0}[\frac{sinx^2.sin^2(\frac{x^2}{2})}{x^6}]$

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

= $2\lim_{x\to0}(\frac{sinx^2}{x^2})(\frac{sin\frac{x^2}{2}}{\frac{x^2}{2}})^2×\frac{1}{4}$

As we know that Lim_x→0[sinx/x] = 1

= 2 × (1/4)

= 1/2

问题18. Lim _x→0 [sin ² (4x ² )/ x ⁴ ]

解决方案：

We have,

Lim_x→0[sin²(4x²)/x⁴]

= Lim_x→0[(sin(4x²))²/(x²)²]

= Lim_x→0[sin(4x²)/4x²]² × 4²

As we know that Lim_x→0[sinx/x] = 1

= 4²

= 16

问题19. Lim _x→0 [(xcosx + 2sinx)/(x ² + tanx)]

解决方案：

We have,

Lim_x→0[(xcosx + 2sinx)/(x² + tanx)]

= Lim_x→0[x(cosx + 2sinx/x)/x(x + tanx/x)]

= Lim_x→0[(cosx + 2sinx/x)/(x + tanx/x)]

=$\frac{\lim_{x\to0}cosx+2\lim_{x\to0}\frac{sinx}{x}}{\lim_{x\to0}x+\lim_{x\to0}\frac{tanx}{x}}$

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

= (1 + 2)/(0 + 1)

= 3

问题20. Lim _x→0 [(2x – sinx)/(x + tanx)]

解决方案：

We have,

Lim_x→0[(2x – sinx)/(x + tanx)]

= Lim_x→0[x(2 – sinx/x)/x(1 + tanx/x)]

= Lim_x→0[(2 – sinx/x)/(1 + tanx/x)]

= $\frac{2-\lim_{x\to0}\frac{sinx}{x}}{1+\lim_{x\to0}\frac{tanx}{x}}$

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

= (2 – 1)/(1 + 1)

= 1/2

问题21. Lim _x→0 [(5xcosx + 3sinx)/(3x ² + tanx)]

解决方案：

We have,

Lim_x→0[(5xcosx + 3sinx)/(3x² + tanx)]

= Lim_x→0[x(5cosx + 3sinx/x)/x(3x + tanx/x)]

= Lim_x→0[(5cosx + 3sinx/x)/(3x + tanx/x)]

= $\frac{5\lim_{x\to0}cosx+3\lim_{x\to0}\frac{sinx}{x}}{3\lim_{x\to0}x+\lim_{x\to0}\frac{tanx}{x}}$

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

= (5 cos0 + 3)/(0 + 1)

= 8

问题22. Lim _x→0 [(sin3x – sinx)/(sinx)]

解决方案：

We have,

Lim_x→0[(sin3x – sinx)/(sinx)]

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

= 2Lim_x→0[cos2x.sinx/sinx]

= 2Lim_x→0cos2x

= 2 × cos0

= 2 × 1

= 2

问题23. Lim _x→0 [(sin5x – sin3x)/(sinx)]

解决方案：

We have,

Lim_x→0[(sin5x – sin3x)/(sinx)]

= $\lim_{x\to0}[\frac{2cos(\frac{5x+3x}{2})sin(\frac{5x-3x}{2})}{sinx}]$

= 2Lim_x→0[cos4x.sinx/sinx]

= 2Lim_x→0cos4x

= 2 × cos0

= 2 × 1

= 2

问题24. Lim _x→0 [(cos3x – cos5x)/ x ² ]

解决方案：

We have,

Lim_x→0[(cos3x – cos5x)/x²]

= $\lim_{x\to0}[\frac{-2sin(\frac{3x+5x}{2})sin(\frac{3x-5x}{2})}{x^2}]$

= Lim_x→0[-2sin4x.sin(-x)/x²]

= Lim_x→0[2sin4x.sinx/x²]

= $2\lim_{x\to0}[\frac{sin4x}{4x}×4]×\lim_{x\to0}[\frac{sinx}{x}]$

As we know that Lim_x→0[sinx/x] = 1 

= 2 × 4 × 1

= 8

问题25. Lim _x→0 [(tan3x – 3x)/(3x – sin ² x)]

解决方案：

We have,

Lim_x→0[(tan3x – 3x)/(3x – sin²x)]

= Lim_x→0[x(tan3x/x – 3)/x(3 – sin²x/x)]

= Lim_x→0[(tan3x/x – 3)/(3 – sin²x/x)]

= $\lim_{x\to0}[\frac{\frac{tan3x}{3x}×3-2}{3-\frac{sinx}{x}×sinx}]$

As we know that Lim_x→0[sinx/x] = 1 

= (3 – 2)/(3 – 0)

= 1/3

问题26. Lim _x→0 [{sin(2 + x)– sin(2 – x)} / x]

解决方案：

We have,

Lim_x→0[{sin(2 + x) – sin(2 – x)}/x]

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

= 2Lim_x→0[cos2.sinx/x]

= 2cos×2Lim_x→0[sinx/x]

As we know that Lim_x→0[sinx/x] = 1 

= 2cos2

问题27.林_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[{a²sin(a + h) + h²sin(a + h) + 2ahsin(a + h) – a²sina}/h]

= $\lim_{h\to0}[a^2(\frac{sin(a+h)-sina}{h}+\frac{h^2sin(a+h)}{h}+\frac{2ahsin(a+h)}{h}]$

= $\lim_{h\to0}[a^2\frac{2cos\frac{(a+h+a)}{2}sin\frac{(a+h-a)}{2}}{2×\frac{h}{2}}+hsin(a+h)+2asin(a+h)]$

= $\lim_{h\to0}[a^2cos(\frac{a+h+a}{2})+hsin(a+h)+2asin(a+h)$

= a²cosa + 0 + 2asina

= a²cosa + 2asina

问题28.林_x→0 [{tanx – sinx} / {sin3x – 3sinx}]

解决方案：

We have,

Lim_x→0[{tanx – sinx}/{sin3x – 3sinx}]

= $\lim_{x\to0}[\frac{\frac{sinx}{cosx}-sinx}{3sinx-4sin^3x-3sinx}]$

= $\frac{-1}{4}\lim_{x\to0}[\frac{sinx(\frac{1}{cosx}-1)}{sin^2x.sinx}]$

= $\frac{-1}{4}\lim_{x\to0}[\frac{(\frac{1-cosx}{cosx})}{1-cos^2x}]$

= $\frac{-1}{4}\lim_{x\to0}[\frac{(1-cosx)}{cosx(1-cosx)(1+cosx)}]$

= $\frac{-1}{4}\lim_{x\to0}[\frac{1}{cosx(1+cosx)}]$

= (-1/4)(1/2)

= -(1/8)

问题29. Lim _x→0 [{sex5x – sec5x} / {sec3x – secx}]

解决方案：

We have,

Lim_x→0[{sex5x – sec5x}/{sec3x – secx}]

= $\lim_{x\to0}\frac{\frac{1}{cos5x}-\frac{1}{cos3x}}{\frac{1}{cos3x}-\frac{1}{cosx}}$

= $\lim_{x\to0}(\frac{cos3x-cos5x}{cosx-cos3x})(\frac{cosx.cos3x}{cos3x.cos5x})$

= $\lim_{x\to0}[\frac{-2sin4x.sin(-x)}{-2sin2x.sin(-x)}][\frac{cosx}{cos5x}]$

= $\lim_{x\to0}[\frac{sin4x}{sin2x}][\frac{cosx}{cos5x}]$

= $\lim_{x\to0}[\frac{\frac{sin4x}{4x}×4x}{\frac{sin2x}{2x}×2x}][\frac{cosx}{cos5x}]$

As we know that Lim_x→0[sinx/x] = 1 

= (4x/2x)(cos0/cos0)

= 2

问题30.林_x→0 [{1 – cos2x} / {cos2x – cos3x}]

解决方案：

We have,

Lim_x→0[{1 – cos2x}/{cos2x – cos3x}]

= $\lim_{x\to0}[\frac{2sin^2x}{-2sin(\frac{3x+7x}{2})sin(\frac{3x-7x}{2})}]$

= $\lim_{x\to0}[\frac{2sin^2x}{2sin(5x)sin(3x)}]$

= $\lim_{x\to0}[\frac{sinx.sinx}{sin(5x)sin(3x)}]$

= $\lim_{x\to0}[\frac{(\frac{sinx}{x}×x)(\frac{sinx}{x}×x)}{(\frac{sin5x}{5x}×5x)(\frac{sin3x}{3x}×3x)}$

As we know that Lim_x→0[sinx/x] = 1 

= (1/5 × 3)

= 1/15

问题31. Lim _x→0 [(1-cos2x + tan ² x)/(xsinx)]

解决方案：

We have,

Lim_x→0[(1 – cos2x + tan²x)/(xsinx)]

= Lim_x→0[(2sin²x + tan²x)/(xsinx)]

Dividing numerator and denominator by x²

= $\lim_{x\to0}\frac{2(\frac{sinx}{x})^2 \times x^2+(\frac{tanx}{x})^2\times x^2}{\lim_{x\to0}(\frac{sinx}{x})\times x^2}$

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

= (2 × 1 × x² + 1) + (1 × x²) /(1 × x²)

= 3