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

Question 1. lim_x→π/2[π/2 – x].tanx

Solution:

We have,

lim_x→π/2[π/2 – x].tanx

Let us considered, y = [π/2 – x]

Here, x→π/2, y→0

= lim_y→0[y.tan(π/2 – y)]

= lim_y→0[y.{sin(π/2 – y)/cos(π/2 – y)]

= lim_y→0[y.{cosy/siny}]

= lim_y→0[y/siny].cosy

= lim_y→0[cosy]      [Since, lim_y→0[siny/y] = 1]

= 1

Question 2. lim_x→π/2[sin2x/cosx]

Solution:

We have,

lim_x→π/2[sin2x/cosx]

= lim_x→π/2[2sinx.cosx/cosx]

= 2Lim_x→π/2[sinx]

= 2

Question 3. lim_x→π/2[cos²x/(1 – sinx)]

Solution:

We have,

lim_x→π/2[cos²x/(1 – sinx)]

= lim_x→π/2[(1 – sin²x)/(1 – sinx)]

= lim_x→π/2[(1 – sinx)(1 + sinx)/(1 – sinx)]

= lim_x→π/2[(1 + sinx)]

= 1 + 1

= 2

Question 4. lim_x→π/2[(1 – sinx)/cos²x]

Solution:

We have,

lim_x→π/2[(1 – sinx)/cos²x]

= lim_x→π/2[(1 – sinx)/(1 – sin²x)]

= lim_x→π/2[(1 – sinx)/(1 – sinx)(1 + sinx)]

= lim_x→π/2[1/(1 + sinx)]

= 1/(1 + 1)

= 1/2

Question 5. lim_x→a[(cosx – cosa)/(x – a)]

Solution:

We have,

lim_x→a[(cosx – cosa)/(x – a)]

= 

=

= -2sin[(a + a)/2] × 1 × (1/2)

= -sina

Question 6. lim_x→π/4[(1 – tanx)/(x – π/4)]

Solution:

We have,

lim_x→π/4[(1 – tanx)/(x – π/4)]

Let us considered, y = [x – π/4]

Here, x→π/4, y→0

= 

= 

= 

= 

= 

= -2 × 1 × [1/(1 – 0)]

= -2

Question 7. lim_x→π/2[(1 – sinx)/(π/4 – x)²]

Solution:

We have,

lim_x→π/2[(1 – sinx)/(π/4 – x)²]

Let us considered, y = [π/2 – x]

Here, x→π/2, y→0

= 

= lim_y→0[(1 – cosy)/y²]

= 

= 2 × 1 × (1/4)

= (1/2)

Question 8. lim_x→π/3[(√3 – tanx)/(π – 3x)]

Solution:

We have,

lim_x→π/3[(√3 – tanx)/(π – 3x)]

Let us considered, y = [π/3 – x]

When, x→π/3, y→0

= 

= 

= 

= 

= 

= (4/3) × 1 × [1/(1 + 0)]

= (4/3)

Question 9. lim_x→a[(asinx – xsina)/(ax² – xa²)]

Solution:

We have,

lim_x→a[(asinx – xsina)/(ax² – xa²)]

= lim_x→a[(asinx – xsina)/{ax(x – a)}]

Let us considered, y = [x – a]

When, x→a, y→0

= lim_y→0[{asin(y + a) – (y + a)sina)}/{a(y + a)y}]

= lim_y→0[(a.siny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]

= lim_y→0[{a.siny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]

= lim_y→0[{a.siny.cosa + a.sina.2sin²(y/2) – t.sina}/{a(y + a)y}]

= lim_y→0[a.siny.cosa/a(y + a)y] – lim_y→0[2.a.sina.sin²(y/2)/a(y + a)y] + lim_y→0[y.sina/a(a + y)y]

= [(a.cosa)/a²] – [(sina)/a²] + 0

= [(a.cosa – sina)/a²]

Question 10. lim_x→π/2[{√2 – √(1 + sinx)}/cos²x]

Solution:

We have,

lim_x→π/2[{√2 – √(1 + sinx)}/cos²x]

On rationalizing the numerator, we get

= lim_x→π/2[{2 – (1 + sinx)}/cos²x{√2 + √(1 + sinx)}]

= lim_x→π/2[(1 – sinx)/(1 – sin²x){√2 + √(1 + sinx)}]

= lim_x→π/2[(1 – sinx)/(1 – sinx)(1 + sinx){√2 + √(1 + sinx)}]

= lim_x→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]

= 1/{(1 + 1)(√2 + √2)}

= 1/4√2

Question 11. lim_x→π/2[{√(2 – sinx) – 1}/(π/2 – x)²]

Solution:

We have,

lim_x→π/2[{√(2 – sinx) – 1}/(π/2 – x)²]

Let us considered, y = [π/2 – x]

Here, x→π/2, y→0

=

= lim_y→0[{√(2 – cosy) – 1}/y²]

On rationalizing the numerator, we get

= lim_y→0[{(2 – cosy) – 1}/y²{√(2 – cosy) – 1}]

= lim_y→0[{1 – cosy}/y²{√(2 – cosy) – 1}]

= 

= 

= 2/4(1 + 1)

= 1/4

Question 12. lim_x→π/4[(√2 – cosx – sinx)/(π/4 – x)²]

Solution:

We have,

lim_x→π/4[(√2 – cosx – sinx)/(π/4 – x)²]

= 

= 

= 

= 

= 2√2/4

= (1/√2)

Question 13. lim_x→π/8[(cot4x – cos4x)/(π – 8x)³]

Solution:

We have,

lim_x→π/8[(cot4x – cos4x)/(π – 8x)³]

= lim_x→π/8[(cot4x – cos4x)/8³(π/8 – x)³]

Let us considered, (π/8 – x) = y

When x→π/8, y→0

=

= lim_x→0[(tan4x-sin4x)/8³(π/8-x)³]

= lim_x→0[(sin4x/cos4x-sin4x)/8³(π/8-x)³]

=

=

=

=

=

= (2 × 4 × 1 × 4 × 1)/(8³)

= 1/16

Question 14. lim_x→a[(cosx – cosa)/(√x – √a)]

Solution:

We have,

lim_x→a[(cosx – cosa)/(√x – √a)]

= 

On rationalizing the denominator, we get

= 

= 

= -2 × sina × 1 × (1/2) × 2√a

= -2√a.sina

Question 15. lim_x→π[{√(5 + cosx) – 2}/(π – x)²]

Solution:

We have,

lim_x→π[{√(5 + cosx) – 2}/(π – x)²]

Let us considered, y = [π – x]

When, x→π, y→0

= 

= lim_y→0[{√(5 – cosy) – 2}/y²]

On rationalizing the numerator, we get

= 

= lim_y→0[{1 – cosy}/y²{√(5 – cosy)-2}]

= 

= 

= 2 × (1/4) × {1/(2 + 2)}

= (1/8)

Question 16. lim_x→a[(cos√x – cos√a)/(x – a)]

Solution:

We have,

lim_x→a[(cos√x – cos√a)/(x – a)]

= 

= 

= -2sin√a × 1 × (1/2√a) × (1/2)

= -(sin√a/2√a)

Question 17. lim_x→a[(sin√x – sin√a)/(x – a)]

Solution:

We have,

lim_x→a[(sin√x – sin√a)/(x – a)]

= 

= 

= 2cos√a × 1 × (1/2√a) × (1/2)

= (cos√a/2√a)

Question 18. lim_x→1[(1 – x²)/sin2πx]

Solution:

We have,

lim_x→1[(1 – x²)/sin2πx]

When, x→1, h→0

= lim_h→0[{1-(1-h)²}/sin2π(1-h)]

= lim_h→0[(2h-h²)/-sin2πh]

= lim_h→0[{h(2-h)}/sin2πh]

=

= 

= -2/2π

= -1/π

Question 19. lim_x→π/4[{f(x) – f(π/4)}/{x – π/4}]

Solution:

We have,

lim_x→π/4[{f(x) – f(π/4)}/{x – π/4}]

When, x→π/4, h→0

= lim_h→0[{f(π/4 + h) – f(π/4)}/{π/4 + h – π/4}]

It is given that f(x) = sin2x

= lim_h→0[{sin(π/2 + 2h) – sin(π/2)}/h]

= lim_h→0[(cos2h – 1)/h]

= lim_h→0[{-2sin²h}/h]

= -2Lim_h→0[(sinh/h)²] × h

= -2 × 1 × 0

= 0