# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.8 | Set 1

Last Updated : 08 May, 2021

### Question 1. limxâ†’Ï€/2[Ï€/2 – x].tanx

Solution:

We have,

limxâ†’Ï€/2[Ï€/2 – x].tanx

Let us considered, y = [Ï€/2 – x]

Here, xâ†’Ï€/2, yâ†’0

= limyâ†’0[y.tan(Ï€/2 – y)]

= limyâ†’0[y.{sin(Ï€/2 – y)/cos(Ï€/2 – y)]

= limyâ†’0[y.{cosy/siny}]

= limyâ†’0[y/siny].cosy

= limyâ†’0[cosy]      [Since, limyâ†’0[siny/y] = 1]

= 1

### Question 2. limxâ†’Ï€/2[sin2x/cosx]

Solution:

We have,

limxâ†’Ï€/2[sin2x/cosx]

= limxâ†’Ï€/2[2sinx.cosx/cosx]

= 2Limxâ†’Ï€/2[sinx]

= 2

### Question 3. limxâ†’Ï€/2[cos2x/(1 – sinx)]

Solution:

We have,

limxâ†’Ï€/2[cos2x/(1 – sinx)]

= limxâ†’Ï€/2[(1 – sin2x)/(1 – sinx)]

= limxâ†’Ï€/2[(1 – sinx)(1 + sinx)/(1 – sinx)]

= limxâ†’Ï€/2[(1 + sinx)]

= 1 + 1

= 2

### Question 4. limxâ†’Ï€/2[(1 – sinx)/cos2x]

Solution:

We have,

limxâ†’Ï€/2[(1 – sinx)/cos2x]

= limxâ†’Ï€/2[(1 – sinx)/(1 – sin2x)]

= limxâ†’Ï€/2[(1 – sinx)/(1 – sinx)(1 + sinx)]

= limxâ†’Ï€/2[1/(1 + sinx)]

= 1/(1 + 1)

= 1/2

### Question 5. limxâ†’a[(cosx – cosa)/(x – a)]

Solution:

We have,

limxâ†’a[(cosx – cosa)/(x – a)]

=

= -2sin[(a + a)/2] Ã— 1 Ã— (1/2)

= -sina

### Question 6. limxâ†’Ï€/4[(1 – tanx)/(x – Ï€/4)]

Solution:

We have,

limxâ†’Ï€/4[(1 – tanx)/(x – Ï€/4)]

Let us considered, y = [x – Ï€/4]

Here, xâ†’Ï€/4, yâ†’0

= -2 Ã— 1 Ã— [1/(1 – 0)]

= -2

### Question 7. limxâ†’Ï€/2[(1 – sinx)/(Ï€/4 – x)2]

Solution:

We have,

limxâ†’Ï€/2[(1 – sinx)/(Ï€/4 – x)2]

Let us considered, y = [Ï€/2 – x]

Here, xâ†’Ï€/2, yâ†’0

= limyâ†’0[(1 – cosy)/y2]

= 2 Ã— 1 Ã— (1/4)

= (1/2)

### Question 8. limxâ†’Ï€/3[(âˆš3 – tanx)/(Ï€ – 3x)]

Solution:

We have,

limxâ†’Ï€/3[(âˆš3 – tanx)/(Ï€ – 3x)]

Let us considered, y = [Ï€/3 – x]

When, xâ†’Ï€/3, yâ†’0

= (4/3) Ã— 1 Ã— [1/(1 + 0)]

= (4/3)

### Question 9. limxâ†’a[(asinx – xsina)/(ax2 – xa2)]

Solution:

We have,

limxâ†’a[(asinx – xsina)/(ax2 – xa2)]

= limxâ†’a[(asinx – xsina)/{ax(x – a)}]

Let us considered, y = [x – a]

When, xâ†’a, yâ†’0

= limyâ†’0[{asin(y + a) – (y + a)sina)}/{a(y + a)y}]

= limyâ†’0[(a.siny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]

= limyâ†’0[{a.siny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]

= limyâ†’0[{a.siny.cosa + a.sina.2sin2(y/2) – t.sina}/{a(y + a)y}]

= limyâ†’0[a.siny.cosa/a(y + a)y] – limyâ†’0[2.a.sina.sin2(y/2)/a(y + a)y] + limyâ†’0[y.sina/a(a + y)y]

= [(a.cosa)/a2] – [(sina)/a2] + 0

= [(a.cosa – sina)/a2]

### Question 10. limxâ†’Ï€/2[{âˆš2 – âˆš(1 + sinx)}/cos2x]

Solution:

We have,

limxâ†’Ï€/2[{âˆš2 – âˆš(1 + sinx)}/cos2x]

On rationalizing the numerator, we get

= limxâ†’Ï€/2[{2 – (1 + sinx)}/cos2x{âˆš2 + âˆš(1 + sinx)}]

= limxâ†’Ï€/2[(1 – sinx)/(1 – sin2x){âˆš2 + âˆš(1 + sinx)}]

= limxâ†’Ï€/2[(1 – sinx)/(1 – sinx)(1 + sinx){âˆš2 + âˆš(1 + sinx)}]

= limxâ†’Ï€/2[1/(1 + sinx){âˆš2 + âˆš(1 + sinx)}]

= 1/{(1 + 1)(âˆš2 + âˆš2)}

= 1/4âˆš2

### Question 11. limxâ†’Ï€/2[{âˆš(2 – sinx) – 1}/(Ï€/2 – x)2]

Solution:

We have,

limxâ†’Ï€/2[{âˆš(2 – sinx) – 1}/(Ï€/2 – x)2]

Let us considered, y = [Ï€/2 – x]

Here, xâ†’Ï€/2, yâ†’0

=

= limyâ†’0[{âˆš(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

= limyâ†’0[{(2 – cosy) – 1}/y2{âˆš(2 – cosy) – 1}]

= limyâ†’0[{1 – cosy}/y2{âˆš(2 – cosy) – 1}]

= 2/4(1 + 1)

= 1/4

### Question 12. limxâ†’Ï€/4[(âˆš2 – cosx – sinx)/(Ï€/4 – x)2]

Solution:

We have,

limxâ†’Ï€/4[(âˆš2 – cosx – sinx)/(Ï€/4 – x)2]

= 2âˆš2/4

= (1/âˆš2)

### Question 13. limxâ†’Ï€/8[(cot4x – cos4x)/(Ï€ – 8x)3]

Solution:

We have,

limxâ†’Ï€/8[(cot4x – cos4x)/(Ï€ – 8x)3]

= limxâ†’Ï€/8[(cot4x – cos4x)/83(Ï€/8 – x)3]

Let us considered, (Ï€/8 – x) = y

When xâ†’Ï€/8, yâ†’0

=

= limxâ†’0[(tan4x-sin4x)/83(Ï€/8-x)3]

= limxâ†’0[(sin4x/cos4x-sin4x)/83(Ï€/8-x)3]

=

=

=

=

=

= (2 Ã— 4 Ã— 1 Ã— 4 Ã— 1)/(83)

= 1/16

### Question 14. limxâ†’a[(cosx – cosa)/(âˆšx – âˆša)]

Solution:

We have,

limxâ†’a[(cosx – cosa)/(âˆšx – âˆša)]

On rationalizing the denominator, we get

= -2 Ã— sina Ã— 1 Ã— (1/2) Ã— 2âˆša

= -2âˆša.sina

### Question 15. limxâ†’Ï€[{âˆš(5 + cosx) – 2}/(Ï€ – x)2]

Solution:

We have,

limxâ†’Ï€[{âˆš(5 + cosx) – 2}/(Ï€ – x)2]

Let us considered, y = [Ï€ – x]

When, xâ†’Ï€, yâ†’0

= limyâ†’0[{âˆš(5 – cosy) – 2}/y2]

On rationalizing the numerator, we get

= limyâ†’0[{1 – cosy}/y2{âˆš(5 – cosy)-2}]

= 2 Ã— (1/4) Ã— {1/(2 + 2)}

= (1/8)

### Question 16. limxâ†’a[(cosâˆšx – cosâˆša)/(x – a)]

Solution:

We have,

limxâ†’a[(cosâˆšx – cosâˆša)/(x – a)]

= -2sinâˆša Ã— 1 Ã— (1/2âˆša) Ã— (1/2)

= -(sinâˆša/2âˆša)

### Question 17. limxâ†’a[(sinâˆšx – sinâˆša)/(x – a)]

Solution:

We have,

limxâ†’a[(sinâˆšx – sinâˆša)/(x – a)]

= 2cosâˆša Ã— 1 Ã— (1/2âˆša) Ã— (1/2)

= (cosâˆša/2âˆša)

### Question 18. limxâ†’1[(1 – x2)/sin2Ï€x]

Solution:

We have,

limxâ†’1[(1 – x2)/sin2Ï€x]

When, xâ†’1, hâ†’0

= limhâ†’0[{1-(1-h)2}/sin2Ï€(1-h)]

= limhâ†’0[(2h-h2)/-sin2Ï€h]

= limhâ†’0[{h(2-h)}/sin2Ï€h]

=

= -2/2Ï€

= -1/Ï€

### Question 19. limxâ†’Ï€/4[{f(x) – f(Ï€/4)}/{x – Ï€/4}]

Solution:

We have,

limxâ†’Ï€/4[{f(x) – f(Ï€/4)}/{x – Ï€/4}]

When, xâ†’Ï€/4, hâ†’0

= limhâ†’0[{f(Ï€/4 + h) – f(Ï€/4)}/{Ï€/4 + h – Ï€/4}]

It is given that f(x) = sin2x

= limhâ†’0[{sin(Ï€/2 + 2h) – sin(Ï€/2)}/h]

= limhâ†’0[(cos2h – 1)/h]

= limhâ†’0[{-2sin2h}/h]

= -2Limhâ†’0[(sinh/h)2] Ã— h

= -2 Ã— 1 Ã— 0

= 0

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