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].tanx
Let 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[cos²x/(1 – sinx)]

Solution:

We have,
lim_x→π/2[cos²x/(1 – sinx)]
= lim_x→π/2[(1 – sin²x)/(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)/cos²x]

Solution:

We have,
lim_x→π/2[(1 – sinx)/cos²x]
= lim_x→π/2[(1 – sinx)/(1 – sin²x)]
= 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)]
= $\lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})}{x-a}]$
= $-2\lim_{x\to a}[sin\frac{(x+a)}{2}]\lim_{x\to a}[\frac{sin\frac{(x-a)}{2}}{\frac{(x-a)}{2}×2}]$
= -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
= $\lim_{y\to0}[\frac{1-tan(y+\frac{π}{4})}{y}]$
= $\lim_{y\to0}[\frac{1-\frac{(tany+tan\frac{π}{4})}{(1-tany.tan\frac{π}{4}}}{y}]$
= $\lim_{y\to0}[\frac{(1-tany-1-tany)}{y(1-tany)}]$
= $\lim_{y\to0}[\frac{-2tany}{y(1-tany)}]$
= $-2\lim_{y\to0}[\frac{tany}{y}]×\lim_{y\to0}[\frac{1}{(1-tany)}]$
= -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
= $\lim_{y\to0}[\frac{1-sin(\frac{π}{2}-y)}{y^2}]$
= lim_y→0[(1 – cosy)/y²]
= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2×4}]$
= 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
= $\lim_{y\to0}[\frac{\sqrt3-tan(\frac{π}{3}-y)}{3y}]$
= $\lim_{y\to0}[\frac{\sqrt3-\frac{(tan\frac{π}{3}-tany)}{(1+tany.tan\frac{π}{3}}}{3y}]$
= $\lim_{y\to0}[\frac{\sqrt3-\frac{(\sqrt{3}-tany)}{(1+\sqrt{3}.tany)}}{3y}]$
= $\lim_{y\to0}[\frac{4tany}{3y(1+\sqrt{3}tany)}]$
= $\frac{4}{3}\lim_{y\to0}[\frac{tany}{y}]×\lim_{y\to0}[\frac{1}{(1+\sqrt3tany)}]$
= (4/3) × 1 × [1/(1 + 0)]
= (4/3)

Question 9. lim_x→a[(asinx – xsina)/(ax²– xa²)]

Solution:

We have,
lim_x→a[(asinx – xsina)/(ax²– xa²)]
= 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.2sin²(y/2) – t.sina}/{a(y + a)y}]
= lim_y→0[a.siny.cosa/a(y + a)y] – lim_y→0[2.a.sina.sin²(y/2)/a(y + a)y] + lim_y→0[y.sina/a(a + y)y]
= [(a.cosa)/a²] – [(sina)/a²] + 0
= [(a.cosa – sina)/a²]

Question 10. lim_x→π/2[{√2 – √(1 + sinx)}/cos²x]

Solution:

We have,
lim_x→π/2[{√2 – √(1 + sinx)}/cos²x]
On rationalizing the numerator, we get
= lim_x→π/2[{2 – (1 + sinx)}/cos²x{√2 + √(1 + sinx)}]
= lim_x→π/2[(1 – sinx)/(1 – sin²x){√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
= $\lim_{y\to0}[\frac{\sqrt{2-sin(\frac{π}{2}-y)}-1}{y^2}]$
= lim_y→0[{√(2 – cosy) – 1}/y²]
On rationalizing the numerator, we get
= lim_y→0[{(2 – cosy) – 1}/y²{√(2 – cosy) – 1}]
= lim_y→0[{1 – cosy}/y²{√(2 – cosy) – 1}]
= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{y^2(\sqrt{2-cosx}+1)}]$
= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2(\sqrt{2-cosx}+1)×4}]$
= 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]
= $\lim_{x\to \frac{π}{2}}\frac{\sqrt2[1-(\frac{1}{\sqrt2}.cosx+\frac{1}{\sqrt2}.sinx)]}{(\frac{π}{4}-x)^2}$
= $\lim_{x\to \frac{π}{2}}\frac{\sqrt{2}[1-cos(\frac{π}{4}-x)]}{(\frac{π}{4}-x)^2}$
= $\lim_{x\to \frac{π}{2}}\frac{[2\sqrt{2}sin^2\frac{(\frac{π}{4}-x)}{2}]}{(\frac{π}{4}-x)^2}$
= $2\sqrt{2}\lim_{x\to \frac{π}{2}}\frac{[sin^2\frac{(\frac{π}{4}-x)}{2}]}{\frac{(\frac{π}{4}-x)}{4}^2×4}$
= 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)/8³(π/8 – x)³]
Let us considered, (π/8 – x) = y
When x→π/8, y→0
= $\lim_{x\to0}[\frac{cot(\frac{π}{2}-4x)-cos(\frac{π}{2}-4x)}{(π-8x)^3}]$
= lim_x→0[(tan4x-sin4x)/8³(π/8-x)³]
= lim_x→0[(sin4x/cos4x-sin4x)/8³(π/8-x)³]
= $\lim_{y\to0}[\frac{sin4y-sin4y.cos4y}{cos4y.8^3.y^3}]$
= $\lim_{y\to0}[\frac{sin4y(1-cos4y)}{cos4y.8^3.y^3}]$
= $\lim_{y\to0}[\frac{sin4y(2sin^22y)}{cos4y.8^3.y^3}]$
= $\frac{2}{8^3}\lim_{y\to0}\frac{sin4y}{y}×\lim_{y\to0}×\frac{sin^22y}{y^2}\lim_{y\to0}\frac{1}{cos4y}$
= $\frac{2}{8^3}\lim_{y\to0}\frac{sin4y}{4y}×4×\lim_{y\to0}×\frac{sin^22y}{(2y)^2}×4×\lim_{y\to0}\frac{1}{cos4y}$
= (2 × 4 × 1 × 4 × 1)/(8³)
= 1/16

Question 14. lim_x→a[(cosx – cosa)/(√x – √a)]

Solution:

We have,
lim_x→a[(cosx – cosa)/(√x – √a)]
= $\lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})}{\sqrt{x}-\sqrt{a}}]$
On rationalizing the denominator, we get
= $\lim_{x\to a}[\frac{-2sin(\frac{x+a}{2})sin(\frac{x-a}{2})(\sqrt{x}+\sqrt{a})}{x-a}]$
= $-2\lim_{x\to a}[sin\frac{(x+a)}{2}]\lim_{x\to a}[\frac{sin\frac{(x-a)}{2}}{\frac{(x-a)}{2}×2}]×\lim_{x\to a}(\sqrt{x}+\sqrt{a})$
= -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
= $\lim_{y\to0}[\frac{\sqrt{5+cos(π-y)}-2}{y^2}]$
= lim_y→0[{√(5 – cosy) – 2}/y²]
On rationalizing the numerator, we get
= $\lim_{y\to0}[\frac{5-cosy-4}{y^2(\sqrt{5-cosy}-2)}]$
= lim_y→0[{1 – cosy}/y²{√(5 – cosy)-2}]
= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{y^2(\sqrt{5-cosx}+2)}]$
= $\lim_{y\to0}[\frac{2sin^2\frac{y}{2}}{(\frac{y}{2})^2(\sqrt{5-cosx}+2)×4}]$
= 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)]
= $\lim_{x\to a}[\frac{-2sin(\frac{\sqrt{x}+\sqrt{a}}{2})sin(\frac{\sqrt{x}-\sqrt{a}}{2})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}]$
= $-2\lim_{x\to a}[sin\frac{(\sqrt{x}+\sqrt{a})}{2}]\lim_{x\to a}[\frac{sin\frac{(\sqrt{x}-\sqrt{a})}{2}}{\frac{(\sqrt{x}-\sqrt{a})}{2}×2}]×\lim_{x\to a}\frac{1}{(\sqrt{x}+\sqrt{a})}$
= -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)]
= $\lim_{x\to a}[\frac{2cos(\frac{\sqrt{x}+\sqrt{a}}{2})sin(\frac{\sqrt{x}-\sqrt{a}}{2})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}]$
= $2\lim_{x\to a}[cos\frac{(\sqrt{x}+\sqrt{a})}{2}]\lim_{x\to a}[\frac{sin\frac{(\sqrt{x}-\sqrt{a})}{2}}{\frac{(\sqrt{x}-\sqrt{a})}{2}×2}]×\lim_{x\to a}\frac{1}{(\sqrt{x}+\sqrt{a})}$
= 2cos√a × 1 × (1/2√a) × (1/2)
= (cos√a/2√a)

Question 18. lim_x→1[(1 – x²)/sin2πx]

Solution:

We have,
lim_x→1[(1 – x²)/sin2πx]
When, x→1, h→0
= lim_h→0[{1-(1-h)²}/sin2π(1-h)]
= lim_h→0[(2h-h²)/-sin2πh]
= lim_h→0[{h(2-h)}/sin2πh]
= $-\lim_{h\to0}[\frac{(2-h)}{\frac{sin2πh}{h}}]$
= $-\lim_{h\to0}[\frac{(2-h)}{(\frac{sin2πh}{2πh})×2π}]$
= -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[{-2sin²h}/h]
= -2Lim_h→0[(sinh/h)²] × h
= -2 × 1 × 0
= 0

Last Updated : 08 May, 2021

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Chapter 1: Sets

Chapter 2: Relations

Chapter 3: Functions

Chapter 4: Measurement of Angles

Chapter 5: Trigonometric Functions

Chapter 6: Graphs of Trigonometric Functions

Chapter 7: Trigonometric Ratios of Compound Angles

Chapter 8: Transformation Formulae

Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles

Chapter 10: Sine and Cosine Formulae and their Applications

Chapter 11: Trigonometric Equations

Chapter 12: Mathematical Induction

Chapter 13: Complex Numbers

Chapter 15: Linear Inequations

Chapter 16: Permutations

Chapter 17: Combinations

Chapter 18: Binomial Theorem

Chapter 19: Arithmetic Progressions

Chapter 20: Geometric Progressions

Chapter 21: Some Special Series

Chapter 22: Brief Review of Cartesian System of Rectangular Coordinates

Chapter 23: The Straight Lines

Chapter 24: The Circle

Chapter 25: Parabola

Chapter 26: Ellipse

Chapter 27: Hyperbola

Chapter 28: Introduction to 3D Coordinate Geometry

Chapter 29: Limits

Chapter 30: Derivatives

Chapter 31: Mathematical Reasoning

Chapter 32: Statistics

Chapter 33: Probability

Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.8 | Set 1

Question 1. limx→π/2[π/2 – x].tanx

Question 2. limx→π/2[sin2x/cosx]

Question 3. limx→π/2[cos2x/(1 – sinx)]

Question 4. limx→π/2[(1 – sinx)/cos2x]

Question 5. limx→a[(cosx – cosa)/(x – a)]

Question 6. limx→π/4[(1 – tanx)/(x – π/4)]

Question 7. limx→π/2[(1 – sinx)/(π/4 – x)2]

Question 8. limx→π/3[(√3 – tanx)/(π – 3x)]

Question 9. limx→a[(asinx – xsina)/(ax2 – xa2)]

Question 10. limx→π/2[{√2 – √(1 + sinx)}/cos2x]

Question 11. limx→π/2[{√(2 – sinx) – 1}/(π/2 – x)2]

Question 12. limx→π/4[(√2 – cosx – sinx)/(π/4 – x)2]

Question 13. limx→π/8[(cot4x – cos4x)/(π – 8x)3]

Question 14. limx→a[(cosx – cosa)/(√x – √a)]

Question 15. limx→π[{√(5 + cosx) – 2}/(π – x)2]

Question 1. lim_x→π/2[π/2 – x].tanx

Question 2. lim_x→π/2[sin2x/cosx]

Question 3. lim_x→π/2[cos²x/(1 – sinx)]

Question 4. lim_x→π/2[(1 – sinx)/cos²x]

Question 5. lim_x→a[(cosx – cosa)/(x – a)]

Question 6. lim_x→π/4[(1 – tanx)/(x – π/4)]

Question 7. lim_x→π/2[(1 – sinx)/(π/4 – x)²]

Question 8. lim_x→π/3[(√3 – tanx)/(π – 3x)]

Question 9. lim_x→a[(asinx – xsina)/(ax²– xa²)]

Question 10. lim_x→π/2[{√2 – √(1 + sinx)}/cos²x]

Question 11. lim_x→π/2[{√(2 – sinx) – 1}/(π/2 – x)²]

Question 12. lim_x→π/4[(√2 – cosx – sinx)/(π/4 – x)²]

Question 13. lim_x→π/8[(cot4x – cos4x)/(π – 8x)³]

Question 14. lim_x→a[(cosx – cosa)/(√x – √a)]

Question 15. lim_x→π[{√(5 + cosx) – 2}/(π – x)²]

Question 16. lim_x→a[(cos√x – cos√a)/(x – a)]

Question 17. lim_x→a[(sin√x – sin√a)/(x – a)]

Question 18. lim_x→1[(1 – x²)/sin2πx]

Question 19. lim_x→π/4[{f(x) – f(π/4)}/{x – π/4}]