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

• Last Updated : 08 May, 2021

### Question 20. limxâ†’1[(1 + cosÏ€x)/(1 – x)2]

Solution:

We have,

limxâ†’1[(1 + cosÏ€x)/(1 – x)2]

Here, xâ†’1, hâ†’0

= Limhâ†’0[{1 + cosÏ€(1 + h)}/{1 – (1 + h)}2]

= Limhâ†’0[(1 – cosÏ€h)/h2]

= Limhâ†’0[2sin2(Ï€h/2)/h2]

= 2Ï€2/4

= Ï€2/2

### Question 21. limxâ†’1[(1 – x2)/sinÏ€x]

Solution:

We have,

limxâ†’1[(1 – x2)/sinÏ€x]

Here, xâ†’1, hâ†’0

= limhâ†’0[{1 – (1 – h)2}/sinÏ€(1 – h)]

= limhâ†’0[(2h – h2)/-sinÏ€h]

= -limhâ†’0[{h(2 – h)}/sinÏ€h]

=

=

= (2 – 0)/Ï€

= 2/Ï€

### Question 22. limxâ†’Ï€/4[(1 – sin2x)/(1 + cos4x)]

Solution:

We have,

limxâ†’Ï€/4[(1 – sin2x)/(1 + cos4x)]

Here, xâ†’Ï€/4, hâ†’0

= limhâ†’0[{1 – sin2(Ï€/4 – h)}/{1 + cos4(Ï€/4 – h)}]

= limhâ†’0[{1 – sin(Ï€/2 – 2h)}/{1 + cos(Ï€ – 4h)}]

= limhâ†’0[(1 – cos2h)/(1 – cos4h)]

= limhâ†’0[2sin2h/2sin22h]

=

= (1/4)

### Question 23. limxâ†’Ï€[(1 + cosx)/tan2x]

Solution:

We have,

limxâ†’Ï€[(1 + cosx)/tan2x]

Here, xâ†’Ï€, hâ†’0

= limhâ†’0[{1+cos(Ï€ + h)}/tan2(Ï€ + h)]

= limhâ†’0[(1 – cosh)/tan2h]

= limhâ†’0[{2sin2(h/2)}/tan2h]

= 2/4

= 1/2

### Question 24. limnâ†’âˆž[nsin(Ï€/4n)cos(Ï€/4n)]

Solution:

We have,

limnâ†’âˆž[nsin(Ï€/4n)cos(Ï€/4n)]

= limnâ†’âˆž[nsin(Ï€/4n)]Limnâ†’âˆž[cos(Ï€/4n)]

=

=

Let, y = (Ï€/4n)

If nâ†’âˆž, yâ†’0.

= (Ï€/4).Limyâ†’0[siny/y]

= (Ï€/4)

### Question 25. limnâ†’âˆž[2n-1sin(a/2n)]

Solution:

We have,

limnâ†’âˆž[2n-1sin(a/2n)]

=

=

=

Let, y = (a/2n)

If nâ†’âˆž, yâ†’0

= (a/2).Limyâ†’0[siny/y]

= (a/2)

### Question 26. limnâ†’âˆž[sin(a/2n)/sin(b/2n)]

Solution:

We have,

limnâ†’âˆž[sin(a/2n)/sin(b/2n)]

=

Let, y = (a/2n) and z = (b/2n)

If nâ†’âˆž, yâ†’0 and zâ†’0

=

=

= (a/b)

### Question 27. limxâ†’-1[(x2 – x – 2)/{(x2 + x) + sin(x + 1)}]

Solution:

We have,

limxâ†’-1[(x2 – x – 2)/{(x2 + x) + sin(x + 1)}]

= limxâ†’-1[(x2 – x – 2)/{x(x + 1) + sin(x + 1)}]

= limxâ†’-1[(x – 2)(x + 1)/{x(x + 1) + sin(x + 1)}]

Let, y = x + 1

If xâ†’-1, then yâ†’0

= limyâ†’0[y(y – 3)/{y(y – 1) + siny}]

=

= (0 – 3)/{(0 – 1) + 1}

= -3/0

= âˆž

### Question 28. limxâ†’2[(x2 – x – 2)/{(x2 – 2x) + sin(x – 2)}]

Solution:

We have,

limxâ†’2[(x2 – x – 2)/{(x2 – 2x) + sin(x – 2)}]

= limxâ†’2[{(x – 2)(x + 1)}/{x(x + 1) + sin(x + 1)}]

Let, y = x – 2

If xâ†’2, then yâ†’0

= limyâ†’0[y(y + 3)/{y(y + 2) + siny}]

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

= 3/3

= 1

### Question 29. limxâ†’1[(1 – x)tan(Ï€x/2)]

Solution:

We have,

limxâ†’1[(1 – x)tan(Ï€x/2)]

Here, xâ†’1, hâ†’0

= limhâ†’0[{1 – (1 – h)}tan{Ï€/2(1 – h)}]

= limhâ†’0[htan{Ï€/2-Ï€h/2)}

= limhâ†’0[hcot(Ï€h/2)]

=

=

=

= (2/Ï€)

### Question 30. limxâ†’Ï€/4[(1 – tanx)/(1 – âˆš2sinx)]

Solution:

We have,

limxâ†’Ï€/4[(1 – tanx)/(1 – âˆš2sinx)]

On rationalizing the denominator.

= limxâ†’Ï€/4[{(1 – tanx)(1 – âˆš2sinx)}/(1 – 2sin2x)]

=

=

=

=

=

= 2/1

= 2

### Question 31. limxâ†’Ï€[{âˆš(2 + cosx) – 1}/(Ï€ – x)2]

Solution:

We have,

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

Let, y = [Ï€ – x]

Here, xâ†’Ï€, yâ†’0

=

= limyâ†’0[{âˆš(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

=

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

=

=

= 2 Ã— (1/4) Ã— {1/(1 + 1)}

= (1/4)

### Question 32. limxâ†’Ï€/4[(âˆšcosx – âˆšsinx)/(x – Ï€/4)]

Solution:

We have,

limxâ†’Ï€/4[(âˆšcosx – âˆšsinx)/(x – Ï€/4)]

On rationalizing the numerator, we get

= limxâ†’Ï€/4[(cosx – sinx)/{(âˆšcosx + âˆšsinx)(x – Ï€/4)}]

=

### Question 33. limxâ†’1[(1 – 1/x)/sinÏ€(x – 1)]

Solution:

We have,

limxâ†’1[(1 – 1/x)/sinÏ€(x – 1)]

= limxâ†’1[(x – 1)/x{sinÏ€(x – 1)}]

Let, y = x – 1

If xâ†’1, then yâ†’0

= limyâ†’0[y/{(y + 1)sin(Ï€y)}]

= 1/{(1 + 0) Ã— 1 Ã— Ï€}

= 1/Ï€

### Question 34. limxâ†’Ï€/6[(cot2x – 3)/(cosecx – 2)]

Solution:

We have,

limxâ†’Ï€/6[(cot2x – 3)/(cosecx – 2)]

= limxâ†’Ï€/6[(cosec2x – 1 – 3)/(cosecx – 2)]

= limxâ†’Ï€/6[(cosec2x – 22)/(cosecx – 2)]

= limxâ†’Ï€/6[{(cosecx + 2)(cosecx – 2)}/(cosecx – 2)]

= limxâ†’Ï€/6[(cosecx + 2)]

= cosec(Ï€/6) + 2

= 2 + 2

= 4

### Question 35. limxâ†’Ï€/4[(âˆš2 – cosx – sinx)/(4x – Ï€)2]

Solution:

We have,

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

= limxâ†’Ï€/4[(âˆš2 – cosx – sinx)/{42(Ï€/4 – x)2}]

=

=

=

=

= 2âˆš2/43

= (2âˆš2 Ã— âˆš2)/(43âˆš2)

= 4/(43âˆš2)

= 1/(16âˆš2)

### Question 36. limxâ†’Ï€/2[{(Ï€/2 – x)sinx – 2cosx}/{(Ï€/2 – x) + cotx}]

Solution:

We have,

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

=

= limhâ†’0[(hcosh-2sinh)/(h+tanh)]

(On dividing the numerator and denominator by h)

= (1 – 2)/(1 + 1)

= -1/2

### Question 37. limxâ†’Ï€/4[(cosx – sinx)/{(Ï€/4 – x)(cosx + sinx)}]

Solution:

We have,

limxâ†’Ï€/4[(cosx – sinx)/{(Ï€/4 – x)(cosx + sinx)}]

On dividing the numerator and denominator by âˆš2, we get

= (âˆš2 Ã— âˆš2)/2

= 1

### Question 38. limxâ†’Ï€[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Solution:

We have,

limxâ†’Ï€[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Let, x = Ï€ + h

If xâ†’Ï€, then hâ†’0

=

= 1/âˆš2

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