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

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  • Last Updated : 08 May, 2021
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Question 1. limx→π/2[π/2 – x].tanx

Solution:

We have,

limx→π/2[π/2 – x].tanx

Let us considered, y = [π/2 – x]

Here, x→π/2, y→0

= limy→0[y.tan(π/2 – y)]

= limy→0[y.{sin(π/2 – y)/cos(π/2 – y)]

= limy→0[y.{cosy/siny}]

= limy→0[y/siny].cosy

= limy→0[cosy]      [Since, limy→0[siny/y] = 1]

= 1

Question 2. limx→π/2[sin2x/cosx]

Solution:

We have,

limx→π/2[sin2x/cosx]

= limx→π/2[2sinx.cosx/cosx]

= 2Limx→π/2[sinx]

= 2

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

Solution:

We have,

limx→π/2[cos2x/(1 – sinx)]

= limx→π/2[(1 – sin2x)/(1 – sinx)]

= limx→π/2[(1 – sinx)(1 + sinx)/(1 – sinx)]

= limx→π/2[(1 + sinx)]

= 1 + 1

= 2

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

Solution:

We have,

limx→π/2[(1 – sinx)/cos2x]

= limx→π/2[(1 – sinx)/(1 – sin2x)]

= limx→π/2[(1 – sinx)/(1 – sinx)(1 + sinx)]

= limx→π/2[1/(1 + sinx)]

= 1/(1 + 1)

= 1/2

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

Solution:

We have,

limx→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. limx→π/4[(1 – tanx)/(x – π/4)]

Solution:

We have,

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

Solution:

We have,

limx→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}]

= limy→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. limx→π/3[(√3 – tanx)/(π – 3x)]

Solution:

We have,

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

Solution:

We have,

limx→a[(asinx – xsina)/(ax2 – xa2)]

= limx→a[(asinx – xsina)/{ax(x – a)}]

Let us considered, y = [x – a]

When, x→a, y→0

= limy→0[{asin(y + a) – (y + a)sina)}/{a(y + a)y}]

= limy→0[(a.siny.cosa + asina.cosy – ysina – asina)/{a(y + a)y}]

= limy→0[{a.siny.cosa + a.sina.(cosy – 1) – y.sina}/{a(y + a)y}]

= limy→0[{a.siny.cosa + a.sina.2sin2(y/2) – t.sina}/{a(y + a)y}]

= limy→0[a.siny.cosa/a(y + a)y] – limy→0[2.a.sina.sin2(y/2)/a(y + a)y] + limy→0[y.sina/a(a + y)y]

= [(a.cosa)/a2] – [(sina)/a2] + 0

= [(a.cosa – sina)/a2]

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

Solution:

We have,

limx→π/2[{√2 – √(1 + sinx)}/cos2x]

On rationalizing the numerator, we get

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

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

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

= limx→π/2[1/(1 + sinx){√2 + √(1 + sinx)}]

= 1/{(1 + 1)(√2 + √2)}

= 1/4√2

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

Solution:

We have,

limx→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}]

= limy→0[{√(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

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

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

Solution:

We have,

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

Solution:

We have,

limx→π/8[(cot4x – cos4x)/(π – 8x)3]

= limx→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}]

= limx→0[(tan4x-sin4x)/83(π/8-x)3]

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

Solution:

We have,

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

Solution:

We have,

limx→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}]

= limy→0[{√(5 – cosy) – 2}/y2]

On rationalizing the numerator, we get

= \lim_{y\to0}[\frac{5-cosy-4}{y^2(\sqrt{5-cosy}-2)}]

= limy→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. limx→a[(cos√x – cos√a)/(x – a)]

Solution:

We have,

limx→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. limx→a[(sin√x – sin√a)/(x – a)]

Solution:

We have,

limx→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. limx→1[(1 – x2)/sin2πx]

Solution:

We have,

limx→1[(1 – x2)/sin2πx]

When, x→1, h→0

= limh→0[{1-(1-h)2}/sin2π(1-h)]

= limh→0[(2h-h2)/-sin2πh]

= limh→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. limx→π/4[{f(x) – f(π/4)}/{x – π/4}]

Solution:

We have,

limx→π/4[{f(x) – f(π/4)}/{x – π/4}]

When, x→π/4, h→0

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

It is given that f(x) = sin2x

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

= limh→0[(cos2h – 1)/h]

= limh→0[{-2sin2h}/h]

= -2Limh→0[(sinh/h)2] × h

= -2 × 1 × 0

= 0


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