# 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,

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free classeswhich will definitely help them in making a wise career choice in the future.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}[2sin^{2}(πh/2)/h^{2}]=

= 2π

^{2}/4= π

^{2}/2

### Question 21. lim_{x→1}[(1 – x^{2})/sinπx]

**Solution:**

We have,

lim

_{x→1}[(1 – x^{2})/sinπx]Here,

x→1, h→0= lim

_{h→0}[{1 – (1 – h)^{2}}/sinπ(1 – h)]= lim

_{h→0}[(2h – h^{2})/-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}[2sin^{2}h/2sin^{2}2h]=

= (1/4)

### Question 23. lim_{x→π}[(1 + cosx)/tan^{2}x]

**Solution:**

We have,

lim

_{x→π}[(1 + cosx)/tan^{2}x]Here,

x→π, h→0= lim

_{h→0}[{1+cos(π + h)}/tan^{2}(π + h)]= lim

_{h→0}[(1 – cosh)/tan^{2}h]= lim

_{h→0}[{2sin^{2}(h/2)}/tan^{2}h]=

= 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-1}sin(a/2^{n})]

**Solution:**

We have,

lim

_{n→∞}[2^{n-1}sin(a/2^{n})]=

=

=

Let, y = (a/2

^{n})If n→∞, y→0

= (a/2).Lim

_{y→0}[siny/y]= (a/2)

### Question 26. lim_{n→∞}[sin(a/2^{n})/sin(b/2^{n})]

**Solution:**

We have,

lim

_{n→∞}[sin(a/2^{n})/sin(b/2^{n})]=

Let, y = (a/2

^{n}) and z = (b/2^{n})If n→∞, y→0 and z→0

=

=

= (a/b)

### Question 27. lim_{x→-1}[(x^{2 }– x – 2)/{(x^{2 }+ x) + sin(x + 1)}]

**Solution:**

We have,

lim

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

_{x→-1}[(x^{2 }– 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}[(x^{2 }– x – 2)/{(x^{2 }– 2x) + sin(x – 2)}]

**Solution:**

We have,

lim

_{x→2}[(x^{2 }– x – 2)/{(x^{2 }– 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 – 2sin^{2}x)]=

=

=

=

=

= 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}[(cot^{2}x – 3)/(cosecx – 2)]

**Solution:**

We have,

lim

_{x→π/6}[(cot^{2}x – 3)/(cosecx – 2)]= lim

_{x→π/6}[(cosec^{2}x – 1 – 3)/(cosecx – 2)]= lim

_{x→π/6}[(cosec^{2}x – 2^{2})/(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)/{4^{2}(π/4 – x)^{2}}]=

=

=

=

= 2√2/4

^{3}= (2√2 × √2)/(4

^{3}√2)= 4/(4

^{3}√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