Class 11 RD Sharma solutions - Chapter 29 Limits - Exercise 29.7 | Set 1

Question 1. Lim_x→0[sin3x/5x]

Solution:

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)           

Question 2. Lim_x→0[sinx°/x]

Solution:

We have,

Lim_x→0[sinx°/x]

As we know that x° = [(πx)/180]

= Lim_x→0[sin{(πx)/180}/x]

= 

= (π/180) × 1

= (π/180)

Question 3. Lim_x→0[x²/sinx²]

Solution:

We have,

Lim_x→0[x²/sinx²]

=

= 

As we know that, Lim_x→0[sinx/x] = 1

= 1/1

= 1

Question 4. Lim_x→0[sinx.cosx/3x]

Solution:

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

Question 5. Lim_x→0[(3sinx - 4sin³x)/x]

Solution:

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

Question 6. Lim_x→0[tan8x/sin2x]

Solution:

We have,

Lim_x→0[tan8x/sin2x]

= 

= 

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

= 8/2

= 4         

Question 7. Lim_x→0[tan(mx)/tan(nx)]

Solution:

We have,

Lim_x→0[tan(mx)/tan(nx)]

= 

= 

As we know that Lim_x→0[tanx/x] = 1

= m × 1/n × 1

= m/n           

Question 8. Lim_x→0[sin5x/tan3x]

Solution:

We have,

Lim_x→0[sin5x/tan3x]

=

=

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

= 5/3 × 1

= 5/3

Question 9. Lim_x→0[sin(xⁿ)/(xⁿ)]

Solution:

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         

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

Solution:

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)]

= 

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

= (7 - 3)/(4 + 1)

= 4/5           

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

Solution:

We have,

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

=

=

=

=

= (a² - b²)/(c² - d²)

Question 12. Lim_x→0[(tan²3x)/x²]

Solution:

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

Question 13. Lim_x→0[(1 - cosmx)/x²]

Solution:

We have,

Lim_x→0[(1 - cosmx)/x²]

= 

= 

= 

As we know that Lim_x→0[sinx/x] = 1

= 2 × (m/2)²

= m²/2

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

Solution:

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)]

=

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

= (2/3) × (4/3)

= 8/9

Question 15. Lim_x→0[(cos3x - cos7x)/x²]

Solution:

We have,

=

= Lim_x→0[-2sin5x.sin(-2x)/x²]

= Lim_x→0[2sin5x.sin2x/x²]

= 

As we know that Lim_x→0[sinx/x] = 1

= 2 × 5 × 2

= 20

Question 16. Lim_θ→0[sin3θ/tan2θ]

Solution:

We have,

Lim_θ→0[sin3θ/tan2θ]

= 

= 

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

= 3/2

Question 17. Lim_x→0[sinx²(1 - cosx²)/x⁶]

Solution:

We have,

Lim_x→0[sinx²(1 - cosx²)/x⁶]

= Lim_x→0[sinx²{2sin(x²/2}/x⁶]

= 

= 

= 

As we know that Lim_x→0[sinx/x] = 1

= 2 × (1/4)

= 1/2

Question 18. Lim_x→0[sin²(4x²)/x⁴]

Solution:

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

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

Solution:

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)]

=

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

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

= 3

Question 20. Lim_x→0[(2x - sinx)/(x + tanx)]

Solution:

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)]

= 

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

= (2 - 1)/(1 + 1)

= 1/2

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

Solution:

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)]

= 

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

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

= 8

Question 22. Lim_x→0[(sin3x - sinx)/(sinx)]

Solution:

We have,

Lim_x→0[(sin3x - sinx)/(sinx)]

=

= 2Lim_x→0[cos2x.sinx/sinx]

= 2Lim_x→0cos2x

= 2 × cos0

= 2 × 1

= 2

Question 23.  Lim_x→0[(sin5x - sin3x)/(sinx)]

Solution:

We have,

Lim_x→0[(sin5x - sin3x)/(sinx)]

= 

= 2Lim_x→0[cos4x.sinx/sinx]

= 2Lim_x→0cos4x

= 2 × cos0

= 2 × 1

= 2

Question 24. Lim_x→0[(cos3x - cos5x)/x²]

Solution:

We have,

Lim_x→0[(cos3x - cos5x)/x²]

= 

= Lim_x→0[-2sin4x.sin(-x)/x²]

= Lim_x→0[2sin4x.sinx/x²]

= 

As we know that Lim_x→0[sinx/x] = 1 

= 2 × 4 × 1

= 8

Question 25. Lim_x→0[(tan3x - 3x)/(3x - sin²x)]

Solution:

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)]

= 

As we know that Lim_x→0[sinx/x] = 1 

= (3 - 2)/(3 - 0)

= 1/3

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

Solution:

We have,

Lim_x→0[{sin(2 + x) - sin(2 - x)}/x]

= 

= 2Lim_x→0[cos2.sinx/x]

= 2cos×2Lim_x→0[sinx/x]

As we know that Lim_x→0[sinx/x] = 1 

= 2cos2

Question 27.  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[{a²sin(a + h) + h²sin(a + h) + 2ahsin(a + h) - a²sina}/h]

= 

= 

= 

= a²cosa + 0 + 2asina

= a²cosa + 2asina

Question 28. Lim_x→0[{tanx - sinx}/{sin3x - 3sinx}]

Solution:

We have,

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

= 

= 

= 

= 

= 

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

= -(1/8)

Question 29. Lim_x→0[{sex5x - sec5x}/{sec3x - secx}]

Solution:

We have,

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

= 

= 

= 

= 

= 

As we know that Lim_x→0[sinx/x] = 1 

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

= 2

Question 30. Lim_x→0[{1 - cos2x}/{cos2x - cos3x}]

Solution:

We have,

Lim_x→0[{1 - cos2x}/{cos2x - cos3x}]

= 

= 

= 

= 

As we know that Lim_x→0[sinx/x] = 1 

= (1/5 × 3)

= 1/15

Question 31. Lim_x→0[(1 - cos2x + tan²x)/(xsinx)]

Solution:

We have,

Lim_x→0[(1 - cos2x + tan²x)/(xsinx)]

= Lim_x→0[(2sin²x + tan²x)/(xsinx)]

Dividing numerator and denominator by x²

= 

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