Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.8 | Set 2

Question 20. lim_x→1[(1 + cosπx)/(1 – x)²]

Solution:

We have,

lim_x→1[(1 + cosπx)/(1 – x)²]

Here, x→1, h→0

= Lim_h→0[{1 + cosπ(1 + h)}/{1 – (1 + h)}²]

= Lim_h→0[(1 – cosπh)/h2]

= Lim_h→0[2sin²(πh/2)/h²]

= 

= 2π²/4

= π²/2

Question 21. lim_x→1[(1 – x²)/sinπx]

Solution:

We have,

lim_x→1[(1 – x²)/sinπx]

Here, x→1, h→0

= lim_h→0[{1 – (1 – h)²}/sinπ(1 – h)]

= lim_h→0[(2h – h²)/-sinπh]

= -lim_h→0[{h(2 – h)}/sinπh]

=

=

= (2 – 0)/π

= 2/π

Question 22. lim_x→π/4[(1 – sin2x)/(1 + cos4x)]

Solution:

We have,

lim_x→π/4[(1 – sin2x)/(1 + cos4x)]

Here, x→π/4, h→0

= lim_h→0[{1 – sin2(π/4 – h)}/{1 + cos4(π/4 – h)}]

= lim_h→0[{1 – sin(π/2 – 2h)}/{1 + cos(π – 4h)}]

= lim_h→0[(1 – cos2h)/(1 – cos4h)]

= lim_h→0[2sin²h/2sin²2h]

=

= (1/4)

Question 23. lim_x→π[(1 + cosx)/tan²x]

Solution:

We have,

lim_x→π[(1 + cosx)/tan²x]

Here, x→π, h→0

= lim_h→0[{1+cos(π + h)}/tan²(π + h)]

= lim_h→0[(1 – cosh)/tan²h]

= lim_h→0[{2sin²(h/2)}/tan²h]

= 

= 2/4

= 1/2

Question 24. lim_n→∞[nsin(π/4n)cos(π/4n)]

Solution:

We have,

lim_n→∞[nsin(π/4n)cos(π/4n)]

= lim_n→∞[nsin(π/4n)]Limn→∞[cos(π/4n)]

=

=

Let, y = (π/4n)

If n→∞, y→0.

= (π/4).Lim_y→0[siny/y]

= (π/4)

Question 25. lim_n→∞[2^n-1sin(a/2ⁿ)]

Solution:

We have,

lim_n→∞[2^n-1sin(a/2ⁿ)]

=

=

=

Let, y = (a/2ⁿ)

If n→∞, y→0

= (a/2).Lim_y→0[siny/y]

= (a/2)

Question 26. lim_n→∞[sin(a/2ⁿ)/sin(b/2ⁿ)]

Solution:

We have,

lim_n→∞[sin(a/2ⁿ)/sin(b/2ⁿ)]

=

Let, y = (a/2ⁿ) and z = (b/2ⁿ)

If n→∞, y→0 and z→0

=

=

= (a/b)

Question 27. lim_x→-1[(x² – x – 2)/{(x² + x) + sin(x + 1)}]

Solution:

We have,

lim_x→-1[(x² – x – 2)/{(x² + x) + sin(x + 1)}]

= lim_x→-1[(x² – x – 2)/{x(x + 1) + sin(x + 1)}]

= lim_x→-1[(x – 2)(x + 1)/{x(x + 1) + sin(x + 1)}]

Let, y = x + 1

If x→-1, then y→0

= lim_y→0[y(y – 3)/{y(y – 1) + siny}]

=

= (0 – 3)/{(0 – 1) + 1}

= -3/0

= ∞

Question 28. lim_x→2[(x² – x – 2)/{(x² – 2x) + sin(x – 2)}]

Solution:

We have,

lim_x→2[(x² – x – 2)/{(x² – 2x) + sin(x – 2)}]

= lim_x→2[{(x – 2)(x + 1)}/{x(x + 1) + sin(x + 1)}]

Let, y = x – 2

If x→2, then y→0

= lim_y→0[y(y + 3)/{y(y + 2) + siny}]

= 

= (0 + 3)/{(0 + 1) + 1}

= 3/3

= 1

Question 29. lim_x→1[(1 – x)tan(πx/2)]

Solution:

We have,

 lim_x→1[(1 – x)tan(πx/2)]

Here, x→1, h→0

= lim_h→0[{1 – (1 – h)}tan{π/2(1 – h)}]

= lim_h→0[htan{π/2-πh/2)}

= lim_h→0[hcot(πh/2)]

=

=

=

= (2/π)

Question 30. lim_x→π/4[(1 – tanx)/(1 – √2sinx)]

Solution:

We have,

lim_x→π/4[(1 – tanx)/(1 – √2sinx)]

On rationalizing the denominator.

= lim_x→π/4[{(1 – tanx)(1 – √2sinx)}/(1 – 2sin²x)]

=

=

=

=

=

= 2/1

= 2

Question 31. lim_x→π[{√(2 + cosx) – 1}/(π – x)²]

Solution:

We have,

lim_x→π[{√(2 + cosx) – 1}/(π – x)²]

Let, y = [π – x]

Here, x→π, y→0

=

= lim_y→0[{√(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

=

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

=

=

= 2 × (1/4) × {1/(1 + 1)}

= (1/4)

Question 32. lim_x→π/4[(√cosx – √sinx)/(x – π/4)]

Solution:

We have,

lim_x→π/4[(√cosx – √sinx)/(x – π/4)]

On rationalizing the numerator, we get

= lim_x→π/4[(cosx – sinx)/{(√cosx + √sinx)(x – π/4)}]

= 

= 

= 

=

Question 33. lim_x→1[(1 – 1/x)/sinπ(x – 1)]

Solution:

We have,

lim_x→1[(1 – 1/x)/sinπ(x – 1)]

= lim_x→1[(x – 1)/x{sinπ(x – 1)}]

Let, y = x – 1

If x→1, then y→0

= lim_y→0[y/{(y + 1)sin(πy)}]

= 

= 

= 1/{(1 + 0) × 1 × π}

= 1/π

Question 34. lim_x→π/6[(cot²x – 3)/(cosecx – 2)]

Solution:

We have,

lim_x→π/6[(cot²x – 3)/(cosecx – 2)]

= lim_x→π/6[(cosec²x – 1 – 3)/(cosecx – 2)]

= lim_x→π/6[(cosec²x – 2²)/(cosecx – 2)]

= lim_x→π/6[{(cosecx + 2)(cosecx – 2)}/(cosecx – 2)]

= lim_x→π/6[(cosecx + 2)]

= cosec(π/6) + 2

= 2 + 2

= 4

Question 35. lim_x→π/4[(√2 – cosx – sinx)/(4x – π)²]

Solution:

We have,

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

= lim_x→π/4[(√2 – cosx – sinx)/{4²(π/4 – x)²}]

=

=

=

=

= 2√2/4³

= (2√2 × √2)/(4³√2)

= 4/(4³√2)

= 1/(16√2)

Question 36. lim_x→π/2[{(π/2 – x)sinx – 2cosx}/{(π/2 – x) + cotx}]

Solution:

We have,

 lim_x→π/2[{(π/2 – x)sinx – 2cosx}/{(π/2 – x) + cotx}]

=

= lim_h→0[(hcosh-2sinh)/(h+tanh)]

=    (On dividing the numerator and denominator by h)

= (1 – 2)/(1 + 1)

= -1/2

Question 37. lim_x→π/4[(cosx – sinx)/{(π/4 – x)(cosx + sinx)}]

Solution:

We have,

lim_x→π/4[(cosx – sinx)/{(π/4 – x)(cosx + sinx)}]

On dividing the numerator and denominator by √2, we get

= 

= 

= 

= 

= 

= (√2 × √2)/2

= 1

Question 38. lim_x→π[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Solution:

We have,

lim_x→π[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Let, x = π + h

If x→π, then h→0

= 

=

= 

= 

= 

= 

= 1/√2