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• RD Sharma Class 11 Solutions for Maths

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

### 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]

= (π/180) × 1

= (π/180)

### Question 3. Limx→0[x2/sinx2]

Solution:

We have,

Limx→0[x2/sinx2]

=

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]

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)]

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]

=

=

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)]

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)}]

=

=

=

=

= (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]

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)]

=

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,

=

= Limx→0[-2sin5x.sin(-2x)/x2]

= Limx→0[2sin5x.sin2x/x2]

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]

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]

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)]

=

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)]

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)]

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)]

=

= 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)]

= 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]

= Limx→0[-2sin4x.sin(-x)/x2]

= Limx→0[2sin4x.sinx/x2]

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)]

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]

= 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]

= a2cosa + 0 + 2asina

= a2cosa + 2asina

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

Solution:

We have,

Limx→0[{tanx – sinx}/{sin3x – 3sinx}]

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

= -(1/8)

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

Solution:

We have,

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

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}]

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

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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