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

• Last Updated : 08 May, 2021

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

Solution:

We have,

limx→1[(1 + cosπx)/(1 – x)2]

Here, x→1, h→0

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

= Limh→0[(1 – cosπh)/h2]

= Limh→0[2sin2(πh/2)/h2]

= 2π2/4

= π2/2

Question 21. limx→1[(1 – x2)/sinπx]

Solution:

We have,

limx→1[(1 – x2)/sinπx]

Here, x→1, h→0

= limh→0[{1 – (1 – h)2}/sinπ(1 – h)]

= limh→0[(2h – h2)/-sinπh]

= -limh→0[{h(2 – h)}/sinπh]

=

=

= (2 – 0)/π

= 2/π

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

Solution:

We have,

limx→π/4[(1 – sin2x)/(1 + cos4x)]

Here, x→π/4, h→0

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

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

= limh→0[(1 – cos2h)/(1 – cos4h)]

= limh→0[2sin2h/2sin22h]

=

= (1/4)

Question 23. limx→π[(1 + cosx)/tan2x]

Solution:

We have,

limx→π[(1 + cosx)/tan2x]

Here, x→π, h→0

= limh→0[{1+cos(π + h)}/tan2(π + h)]

= limh→0[(1 – cosh)/tan2h]

= limh→0[{2sin2(h/2)}/tan2h]

= 2/4

= 1/2

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

Solution:

We have,

limn→∞[nsin(π/4n)cos(π/4n)]

= limn→∞[nsin(π/4n)]Limn→∞[cos(π/4n)]

=

=

Let, y = (π/4n)

If n→∞, y→0.

= (π/4).Limy→0[siny/y]

= (π/4)

Question 25. limn→∞[2n-1sin(a/2n)]

Solution:

We have,

limn→∞[2n-1sin(a/2n)]

=

=

=

Let, y = (a/2n)

If n→∞, y→0

= (a/2).Limy→0[siny/y]

= (a/2)

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

Solution:

We have,

limn→∞[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. limx→-1[(x2 – x – 2)/{(x2 + x) + sin(x + 1)}]

Solution:

We have,

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

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

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

Let, y = x + 1

If x→-1, then y→0

= limy→0[y(y – 3)/{y(y – 1) + siny}]

=

= (0 – 3)/{(0 – 1) + 1}

= -3/0

= ∞

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

Solution:

We have,

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

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

Let, y = x – 2

If x→2, then y→0

= limy→0[y(y + 3)/{y(y + 2) + siny}]

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

= 3/3

= 1

Question 29. limx→1[(1 – x)tan(πx/2)]

Solution:

We have,

limx→1[(1 – x)tan(πx/2)]

Here, x→1, h→0

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

= limh→0[htan{π/2-πh/2)}

= limh→0[hcot(πh/2)]

=

=

=

= (2/π)

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

Solution:

We have,

limx→π/4[(1 – tanx)/(1 – √2sinx)]

On rationalizing the denominator.

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

=

=

=

=

=

= 2/1

= 2

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

Solution:

We have,

limx→π[{√(2 + cosx) – 1}/(π – x)2]

Let, y = [π – x]

Here, x→π, y→0

=

= limy→0[{√(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

=

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

=

=

= 2 × (1/4) × {1/(1 + 1)}

= (1/4)

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

Solution:

We have,

limx→π/4[(√cosx – √sinx)/(x – π/4)]

On rationalizing the numerator, we get

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

=

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

Solution:

We have,

limx→1[(1 – 1/x)/sinπ(x – 1)]

= limx→1[(x – 1)/x{sinπ(x – 1)}]

Let, y = x – 1

If x→1, then y→0

= limy→0[y/{(y + 1)sin(πy)}]

= 1/{(1 + 0) × 1 × π}

= 1/π

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

Solution:

We have,

limx→π/6[(cot2x – 3)/(cosecx – 2)]

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

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

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

= limx→π/6[(cosecx + 2)]

= cosec(π/6) + 2

= 2 + 2

= 4

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

Solution:

We have,

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

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

Solution:

We have,

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

=

= limh→0[(hcosh-2sinh)/(h+tanh)]

(On dividing the numerator and denominator by h)

= (1 – 2)/(1 + 1)

= -1/2

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

Solution:

We have,

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

Solution:

We have,

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

Let, x = π + h

If x→π, then h→0

=

= 1/√2

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