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Class 11 RD Sharma solutions – Chapter 29 Limits – Exercise 29.7 | Set 1
  • Last Updated : 04 May, 2021

Question 1. Limx→0[sin3x/5x]

Solution:

We have,

Limx→0[sin3x/5x]

= (1/5)Limx→0[sin3x/3x] × 3

= (3/5)Limx→0[sin3x/3x]



As we know that, Limx→0[sinx/x] = 1

= (3/5)           

Question 2. Limx→0[sinx°/x]

Solution:

We have,

Limx→0[sinx°/x]

As we know that x° = [(πx)/180]

= Limx→0[sin{(πx)/180}/x]

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



= (π/180) × 1

= (π/180)

Question 3. Limx→0[x2/sinx2]

Solution:

We have,

Limx→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, Limx→0[sinx/x] = 1

= 1/1

= 1



Question 4. Limx→0[sinx.cosx/3x]

Solution:

We have,

Limx→0[(sinx.cosx)/3x]

= Limx→0[(sinx.cosx)/ × 3x]

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

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

= (1/3) × 1 × 1      

= 1/3

Question 5. Limx→0[(3sinx – 4sin3x)/x]

Solution:

We have,

Limx→0[(3sinx – 4sin3x)/x]

As we know that 3sinx – 4sin3x =  sin3x

= Limx→0[(sin3x)/3x] × 3

= 3 × Limx→0[(sin3x)/3x]

As we know that, Limx→0[sinx/x] = 1

= 3 × 1

= 3

Question 6. Limx→0[tan8x/sin2x]

Solution:

We have,

Limx→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 Limx→0[sin2x/2x] = 1 and Limx→0[tan8x/8x] = 1

= 8/2

= 4         

Question 7. Limx→0[tan(mx)/tan(nx)]

Solution:

We have,

Limx→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 Limx→0[tanx/x] = 1

= m × 1/n × 1

= m/n           

Question 8. Limx→0[sin5x/tan3x]

Solution:

We have,

Limx→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 Limx→0[sin5x/5x] = 1 and Limx→0[tan3x/3x] = 1

= 5/3 × 1



= 5/3

Question 9. Limx→0[sin(xn)/(xn)]

Solution:

We have,

Limx→0[sin(xn)/(xn)]

It is in the form of 0/0

So, let xn = y

x→0 than y→0

Limy→0[siny/y]

As we know that Limy→0[siny/y] = 1

= 1         

Question 10. Limx→0[(7xcosx – 3sinx)/(4x + tanx)]

Solution:

We have,

Limx→0[(7xcosx – 3sinx)/(4x + tanx)]

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

= Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

= (7 – 3)/(4 + 1)

= 4/5           

Question 11. Limx→0[{cos(ax) – cos(bx)}/{cos(cx) – cos(dx)}]

Solution:

We have,

Limx→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)

Question 12. Limx→0[(tan23x)/x2]

Solution:

We have,

Limx→0[(tan23x)/x2]



= Limx→0[(tan3x)/x]2

= Limx→0[(tan3x)/3x]2 × 9

As we know that Limx→0[tanx/x] = 1

= 1 × 9

= 9

Question 13. Limx→0[(1 – cosmx)/x2]

Solution:

We have,

Limx→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 Limx→0[sinx/x] = 1

= 2 × (m/2)2

= m2/2

Question 14. Limx→0[(3sin2x + 2x)/(3x + 2tan3x)]

Solution:

We have,

Limx→0[(3sin2x + 2x)/(3x + 2tan3x)]

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

= (2/3)Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

= (2/3) × (4/3)

= 8/9

Question 15. Limx→0[(cos3x – cos7x)/x2]

Solution:

We have,

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

= Limx→0[-2sin5x.sin(-2x)/x2]

= Limx→0[2sin5x.sin2x/x2]

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

As we know that Limx→0[sinx/x] = 1

= 2 × 5 × 2

= 20

Question 16. Limθ→0[sin3θ/tan2θ]

Solution:

We have,

Limθ→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

= 3/2

Question 17. Limx→0[sinx2(1 – cosx2)/x6]

Solution:



We have,

Limx→0[sinx2(1 – cosx2)/x6]

= Limx→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 Limx→0[sinx/x] = 1

= 2 × (1/4)

= 1/2

Question 18. Limx→0[sin2(4x2)/x4]

Solution:

We have,

Limx→0[sin2(4x2)/x4]

= Limx→0[(sin(4x2))2/(x2)2]

= Limx→0[sin(4x2)/4x2]2 × 42

As we know that Limx→0[sinx/x] = 1

= 42

= 16

Question 19. Limx→0[(xcosx + 2sinx)/(x2 + tanx)]

Solution:

We have,

Limx→0[(xcosx + 2sinx)/(x2 + tanx)]

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

= Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

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

= 3

Question 20. Limx→0[(2x – sinx)/(x + tanx)]

Solution:

We have,

Limx→0[(2x – sinx)/(x + tanx)]

= Limx→0[x(2 – sinx/x)/x(1 + tanx/x)]



= Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

= (2 – 1)/(1 + 1)

= 1/2

Question 21. Limx→0[(5xcosx + 3sinx)/(3x2  + tanx)]

Solution:

We have,

Limx→0[(5xcosx + 3sinx)/(3x2 + tanx)]

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

= Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1

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

= 8

Question 22. Limx→0[(sin3x – sinx)/(sinx)]

Solution:

We have,

Limx→0[(sin3x – sinx)/(sinx)]

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

= 2Limx→0[cos2x.sinx/sinx]

= 2Limx→0cos2x

= 2 × cos0

= 2 × 1

= 2

Question 23.  Limx→0[(sin5x – sin3x)/(sinx)]

Solution:

We have,

Limx→0[(sin5x – sin3x)/(sinx)]

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

= 2Limx→0[cos4x.sinx/sinx]

= 2Limx→0cos4x

= 2 × cos0



= 2 × 1

= 2

Question 24. Limx→0[(cos3x – cos5x)/x2]

Solution:

We have,

Limx→0[(cos3x – cos5x)/x2]

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

= Limx→0[-2sin4x.sin(-x)/x2]

= Limx→0[2sin4x.sinx/x2]

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

As we know that Limx→0[sinx/x] = 1 

= 2 × 4 × 1

= 8

Question 25. Limx→0[(tan3x – 3x)/(3x – sin2x)]

Solution:

We have,

Limx→0[(tan3x – 3x)/(3x – sin2x)]

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

= Limx→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 Limx→0[sinx/x] = 1 

= (3 – 2)/(3 – 0)

= 1/3

Question 26. Limx→0[{sin(2 + x) – sin(2 – x)}/x]

Solution:

We have,

Limx→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}]

= 2Limx→0[cos2.sinx/x]

= 2cos×2Limx→0[sinx/x]

As we know that Limx→0[sinx/x] = 1 

= 2cos2

Question 27.  Limh→0[{(a + h)2sin(a + h) – a2sina}/h]

Solution:

We have,

Limh→0[{(a + h)2sin(a + h) – a2sina}/h]

= Limh→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

Question 28. Limx→0[{tanx – sinx}/{sin3x – 3sinx}]

Solution:



We have,

Limx→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)

Question 29. Limx→0[{sex5x – sec5x}/{sec3x – secx}]

Solution:

We have,

Limx→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 Limx→0[sinx/x] = 1 

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

= 2



Question 30. Limx→0[{1 – cos2x}/{cos2x – cos3x}]

Solution:

We have,

Limx→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 Limx→0[sinx/x] = 1 

= (1/5 × 3)

= 1/15

Question 31. Limx→0[(1 – cos2x + tan2x)/(xsinx)]

Solution:

We have,

Limx→0[(1 – cos2x + tan2x)/(xsinx)]

= Limx→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 Limx→0[sinx/x] = 1 and Limx→0[tanx/x] = 1 

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

= 3




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