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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