# 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].tanxLet 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}[cos^{2}x/(1 – sinx)]

**Solution:**

We have,

lim

_{x→π/2}[cos^{2}x/(1 – sinx)]= lim

_{x→π/2}[(1 – sin^{2}x)/(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)/cos^{2}x]

**Solution:**

We have,

lim

_{x→π/2}[(1 – sinx)/cos^{2}x]= lim

_{x→π/2}[(1 – sinx)/(1 – sin^{2}x)]= 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)/y^{2}]=

= 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)/(ax^{2 }– xa^{2})]

**Solution:**

We have,

lim

_{x→a}[(asinx – xsina)/(ax^{2 }– xa^{2})]= 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.2sin^{2}(y/2) – t.sina}/{a(y + a)y}]= lim

_{y→0}[a.siny.cosa/a(y + a)y] – lim_{y→0}[2.a.sina.sin^{2}(y/2)/a(y + a)y] + lim_{y→0}[y.sina/a(a + y)y]= [(a.cosa)/a

^{2}] – [(sina)/a^{2}] + 0= [(a.cosa – sina)/a

^{2}]

### Question 10. lim_{x→π/2}[{√2 – √(1 + sinx)}/cos^{2}x]

**Solution:**

We have,

lim

_{x→π/2}[{√2 – √(1 + sinx)}/cos^{2}x]On rationalizing the numerator, we get

= lim

_{x→π/2}[{2 – (1 + sinx)}/cos^{2}x{√2 + √(1 + sinx)}]= lim

_{x→π/2}[(1 – sinx)/(1 – sin^{2}x){√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}/y^{2}]On rationalizing the numerator, we get

= lim

_{y→0}[{(2 – cosy) – 1}/y^{2}{√(2 – cosy) – 1}]= lim

_{y→0}[{1 – cosy}/y^{2}{√(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)/8^{3}(π/8 – x)^{3}]Let us considered, (π/8 – x) = y

When x→π/8, y→0

=

= lim

_{x→0}[(tan4x-sin4x)/8^{3}(π/8-x)^{3}]= lim

_{x→0}[(sin4x/cos4x-sin4x)/8^{3}(π/8-x)^{3}]=

=

=

=

=

= (2 × 4 × 1 × 4 × 1)/(8

^{3})= 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}/y^{2}]On rationalizing the numerator, we get

=

= lim

_{y→0}[{1 – cosy}/y^{2}{√(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 – x^{2})/sin2πx]

**Solution:**

We have,

lim

_{x→1}[(1 – x^{2})/sin2πx]When, x→1, h→0

= lim

_{h→0}[{1-(1-h)^{2}}/sin2π(1-h)]= lim

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

_{h→0}[(sinh/h)^{2}] × h= -2 × 1 × 0

= 0

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