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

Last Updated : 04 May, 2021

### Question 1. Limxâ†’0[sin3x/5x]

Solution:

We have,

Limxâ†’0[sin3x/5x]

= (1/5)Limxâ†’0[sin3x/3x] Ã— 3

= (3/5)Limxâ†’0[sin3x/3x]

As we know that, Limxâ†’0[sinx/x] = 1

= (3/5)

### Question 2. Limxâ†’0[sinxÂ°/x]

Solution:

We have,

Limxâ†’0[sinxÂ°/x]

As we know that xÂ° = [(Ï€x)/180]

= Limxâ†’0[sin{(Ï€x)/180}/x]

= (Ï€/180) Ã— 1

= (Ï€/180)

### Question 3. Limxâ†’0[x2/sinx2]

Solution:

We have,

Limxâ†’0[x2/sinx2]

=

As we know that, Limxâ†’0[sinx/x] = 1

= 1/1

= 1

### Question 4. Limxâ†’0[sinx.cosx/3x]

Solution:

We have,

Limxâ†’0[(sinx.cosx)/3x]

= Limxâ†’0[(sinx.cosx)/ Ã— 3x]

= 1/3 Limxâ†’0[(sinx)/x] Ã— Limxâ†’0[(cosx)]

As we know that  Limxâ†’0[sinx/x] = 1, and Limxâ†’0cos0 = 1

= (1/3) Ã— 1 Ã— 1

= 1/3

### Question 5. Limxâ†’0[(3sinx – 4sin3x)/x]

Solution:

We have,

Limxâ†’0[(3sinx – 4sin3x)/x]

As we know that 3sinx – 4sin3x =  sin3x

= Limxâ†’0[(sin3x)/3x] Ã— 3

= 3 Ã— Limxâ†’0[(sin3x)/3x]

As we know that, Limxâ†’0[sinx/x] = 1

= 3 Ã— 1

= 3

### Question 6. Limxâ†’0[tan8x/sin2x]

Solution:

We have,

Limxâ†’0[tan8x/sin2x]

As we know that Limxâ†’0[sin2x/2x] = 1 and Limxâ†’0[tan8x/8x] = 1

= 8/2

= 4

### Question 7. Limxâ†’0[tan(mx)/tan(nx)]

Solution:

We have,

Limxâ†’0[tan(mx)/tan(nx)]

As we know that Limxâ†’0[tanx/x] = 1

= m Ã— 1/n Ã— 1

= m/n

### Question 8. Limxâ†’0[sin5x/tan3x]

Solution:

We have,

Limxâ†’0[sin5x/tan3x]

=

=

As we know that Limxâ†’0[sin5x/5x] = 1 and Limxâ†’0[tan3x/3x] = 1

= 5/3 Ã— 1

= 5/3

### Question 9. Limxâ†’0[sin(xn)/(xn)]

Solution:

We have,

Limxâ†’0[sin(xn)/(xn)]

It is in the form of 0/0

So, let xn = y

xâ†’0 than yâ†’0

Limyâ†’0[siny/y]

As we know that Limyâ†’0[siny/y] = 1

= 1

### Question 10. Limxâ†’0[(7xcosx – 3sinx)/(4x + tanx)]

Solution:

We have,

Limxâ†’0[(7xcosx – 3sinx)/(4x + tanx)]

= Limxâ†’0[x(7cosx-3sinx/x)/x(4 + tanx/x)]

= Limxâ†’0[(7cosx-3sinx/x)/(4 + tanx/x)]

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

= (7 – 3)/(4 + 1)

= 4/5

### Question 11. Limxâ†’0[{cos(ax) – cos(bx)}/{cos(cx) – cos(dx)}]

Solution:

We have,

Limxâ†’0[{cos(ax) – cos(bx)}/{cos(cx) – cos(dx)}]

=

=

=

=

= (a2 – b2)/(c2 – d2)

### Question 12. Limxâ†’0[(tan23x)/x2]

Solution:

We have,

Limxâ†’0[(tan23x)/x2]

= Limxâ†’0[(tan3x)/x]2

= Limxâ†’0[(tan3x)/3x]2 Ã— 9

As we know that Limxâ†’0[tanx/x] = 1

= 1 Ã— 9

= 9

### Question 13. Limxâ†’0[(1 – cosmx)/x2]

Solution:

We have,

Limxâ†’0[(1 – cosmx)/x2]

As we know that Limxâ†’0[sinx/x] = 1

= 2 Ã— (m/2)2

= m2/2

### Question 14. Limxâ†’0[(3sin2x + 2x)/(3x + 2tan3x)]

Solution:

We have,

Limxâ†’0[(3sin2x + 2x)/(3x + 2tan3x)]

= Limxâ†’0[2x(1 + 3sin2x/2x)/3 x (1 + 2tan3x/3x)]

= (2/3)Limxâ†’0[(1 + 3sin2x/2x)/3 x (1 + 2tan3x/3x)]

=

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

= (2/3) Ã— (4/3)

= 8/9

### Question 15. Limxâ†’0[(cos3x – cos7x)/x2]

Solution:

We have,

=

= Limxâ†’0[-2sin5x.sin(-2x)/x2]

= Limxâ†’0[2sin5x.sin2x/x2]

As we know that Limxâ†’0[sinx/x] = 1

= 2 Ã— 5 Ã— 2

= 20

### Question 16. LimÎ¸â†’0[sin3Î¸/tan2Î¸]

Solution:

We have,

LimÎ¸â†’0[sin3Î¸/tan2Î¸]

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

= 3/2

### Question 17. Limxâ†’0[sinx2(1 – cosx2)/x6]

Solution:

We have,

Limxâ†’0[sinx2(1 – cosx2)/x6]

= Limxâ†’0[sinx2{2sin(x2/2}/x6]

As we know that Limxâ†’0[sinx/x] = 1

= 2 Ã— (1/4)

= 1/2

### Question 18. Limxâ†’0[sin2(4x2)/x4]

Solution:

We have,

Limxâ†’0[sin2(4x2)/x4]

= Limxâ†’0[(sin(4x2))2/(x2)2]

= Limxâ†’0[sin(4x2)/4x2]2 Ã— 42

As we know that Limxâ†’0[sinx/x] = 1

= 42

= 16

### Question 19. Limxâ†’0[(xcosx + 2sinx)/(x2 + tanx)]

Solution:

We have,

Limxâ†’0[(xcosx + 2sinx)/(x2 + tanx)]

= Limxâ†’0[x(cosx + 2sinx/x)/x(x + tanx/x)]

= Limxâ†’0[(cosx + 2sinx/x)/(x + tanx/x)]

=

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

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

= 3

### Question 20. Limxâ†’0[(2x – sinx)/(x + tanx)]

Solution:

We have,

Limxâ†’0[(2x – sinx)/(x + tanx)]

= Limxâ†’0[x(2 – sinx/x)/x(1 + tanx/x)]

= Limxâ†’0[(2 – sinx/x)/(1 + tanx/x)]

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

= (2 – 1)/(1 + 1)

= 1/2

### Question 21. Limxâ†’0[(5xcosx + 3sinx)/(3x2  + tanx)]

Solution:

We have,

Limxâ†’0[(5xcosx + 3sinx)/(3x2 + tanx)]

= Limxâ†’0[x(5cosx + 3sinx/x)/x(3x + tanx/x)]

= Limxâ†’0[(5cosx + 3sinx/x)/(3x + tanx/x)]

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

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

= 8

### Question 22. Limxâ†’0[(sin3x – sinx)/(sinx)]

Solution:

We have,

Limxâ†’0[(sin3x – sinx)/(sinx)]

=

= 2Limxâ†’0[cos2x.sinx/sinx]

= 2Limxâ†’0cos2x

= 2 Ã— cos0

= 2 Ã— 1

= 2

### Question 23.  Limxâ†’0[(sin5x – sin3x)/(sinx)]

Solution:

We have,

Limxâ†’0[(sin5x – sin3x)/(sinx)]

= 2Limxâ†’0[cos4x.sinx/sinx]

= 2Limxâ†’0cos4x

= 2 Ã— cos0

= 2 Ã— 1

= 2

### Question 24. Limxâ†’0[(cos3x – cos5x)/x2]

Solution:

We have,

Limxâ†’0[(cos3x – cos5x)/x2]

= Limxâ†’0[-2sin4x.sin(-x)/x2]

= Limxâ†’0[2sin4x.sinx/x2]

As we know that Limxâ†’0[sinx/x] = 1

= 2 Ã— 4 Ã— 1

= 8

### Question 25. Limxâ†’0[(tan3x – 3x)/(3x – sin2x)]

Solution:

We have,

Limxâ†’0[(tan3x – 3x)/(3x – sin2x)]

= Limxâ†’0[x(tan3x/x – 3)/x(3 – sin2x/x)]

= Limxâ†’0[(tan3x/x – 3)/(3 – sin2x/x)]

As we know that Limxâ†’0[sinx/x] = 1

= (3 – 2)/(3 – 0)

= 1/3

### Question 26. Limxâ†’0[{sin(2 + x) – sin(2 – x)}/x]

Solution:

We have,

Limxâ†’0[{sin(2 + x) – sin(2 – x)}/x]

= 2Limxâ†’0[cos2.sinx/x]

= 2cosÃ—2Limxâ†’0[sinx/x]

As we know that Limxâ†’0[sinx/x] = 1

= 2cos2

### Question 27.  Limhâ†’0[{(a + h)2sin(a + h) – a2sina}/h]

Solution:

We have,

Limhâ†’0[{(a + h)2sin(a + h) – a2sina}/h]

= Limhâ†’0[{a2sin(a + h) + h2sin(a + h) + 2ahsin(a + h) – a2sina}/h]

= a2cosa + 0 + 2asina

= a2cosa + 2asina

### Question 28. Limxâ†’0[{tanx – sinx}/{sin3x – 3sinx}]

Solution:

We have,

Limxâ†’0[{tanx – sinx}/{sin3x – 3sinx}]

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

= -(1/8)

### Question 29. Limxâ†’0[{sex5x – sec5x}/{sec3x – secx}]

Solution:

We have,

Limxâ†’0[{sex5x – sec5x}/{sec3x – secx}]

As we know that Limxâ†’0[sinx/x] = 1

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

= 2

### Question 30. Limxâ†’0[{1 – cos2x}/{cos2x – cos3x}]

Solution:

We have,

Limxâ†’0[{1 – cos2x}/{cos2x – cos3x}]

As we know that Limxâ†’0[sinx/x] = 1

= (1/5 Ã— 3)

= 1/15

### Question 31. Limxâ†’0[(1 – cos2x + tan2x)/(xsinx)]

Solution:

We have,

Limxâ†’0[(1 – cos2x + tan2x)/(xsinx)]

= Limxâ†’0[(2sin2x + tan2x)/(xsinx)]

Dividing numerator and denominator by x2

As we know that Limxâ†’0[sinx/x] = 1 and Limxâ†’0[tanx/x] = 1

= (2 Ã— 1 Ã— x2 + 1) + (1 Ã— x2) /(1 Ã— x2)

= 3

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