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

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  • Last Updated : 08 May, 2021
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Question 20. limx→1[(1 + cosπx)/(1 – x)2]

Solution:

We have,

limx→1[(1 + cosπx)/(1 – x)2]

Here, x→1, h→0

= Limh→0[{1 + cosπ(1 + h)}/{1 – (1 + h)}2]

= Limh→0[(1 – cosπh)/h2]

= Limh→0[2sin2(πh/2)/h2]

2\lim_{h\to0}[\frac{sin^2(\frac{πh}{2})}{(\frac{2}{π})^2×(\frac{πh}{2})^2}]

= 2π2/4

= π2/2

Question 21. limx→1[(1 – x2)/sinπx]

Solution:

We have,

limx→1[(1 – x2)/sinπx]

Here, x→1, h→0

= limh→0[{1 – (1 – h)2}/sinπ(1 – h)]

= limh→0[(2h – h2)/-sinπh]

= -limh→0[{h(2 – h)}/sinπh]

=\lim_{h\to0}[\frac{(2-h)}{\frac{sinπh}{h}}]

=\lim_{h\to0}[\frac{(2-h)}{(\frac{sinπh}{πh})×π}]

= (2 – 0)/π

= 2/π

Question 22. limx→π/4[(1 – sin2x)/(1 + cos4x)]

Solution:

We have,

limx→π/4[(1 – sin2x)/(1 + cos4x)]

Here, x→π/4, h→0

= limh→0[{1 – sin2(π/4 – h)}/{1 + cos4(π/4 – h)}]

= limh→0[{1 – sin(π/2 – 2h)}/{1 + cos(π – 4h)}]

= limh→0[(1 – cos2h)/(1 – cos4h)]

= limh→0[2sin2h/2sin22h]

=\lim_{h\to0}[\frac{\frac{sin^2h}{h^2}}{\frac{sin^22h}{4h^2}×4}]

= (1/4)

Question 23. limx→π[(1 + cosx)/tan2x]

Solution:

We have,

limx→π[(1 + cosx)/tan2x]

Here, x→π, h→0

= limh→0[{1+cos(π + h)}/tan2(π + h)]

= limh→0[(1 – cosh)/tan2h]

= limh→0[{2sin2(h/2)}/tan2h]

2×\lim_{h\to0}[\frac{sin^2(\frac{h}{2})}{(\frac{h}{2})^2×(\frac{2}{h})^2}]×\lim_{h\to0}[\frac{1}{\frac{tan^2h}{h^2}×h^2}]

= 2/4

= 1/2

Question 24. limn→∞[nsin(π/4n)cos(π/4n)]

Solution:

We have,

limn→∞[nsin(π/4n)cos(π/4n)]

= limn→∞[nsin(π/4n)]Limn→∞[cos(π/4n)]

=\lim_{n\to∞}[\frac{nsin\frac{π}{4n}}{(\frac{π}{4n})}]×(\frac{π}{4n})×1

=(\frac{π}{4})\lim_{n\to∞}[\frac{sin\frac{π}{4n}}{(\frac{π}{4n})}]

Let, y = (π/4n)

If n→∞, y→0.

= (π/4).Limy→0[siny/y]

= (π/4)

Question 25. limn→∞[2n-1sin(a/2n)]

Solution:

We have,

limn→∞[2n-1sin(a/2n)]

=\lim_{n\to∞}[\frac{2^n}{2}×sin(\frac{a}{2^n})]

=\lim_{n\to∞}[\frac{2^n}{2}×\frac{sin(\frac{a}{2^n})}{\frac{a}{2^n}}×\frac{a}{2^n}]

=(\frac{a}{2})\lim_{n\to∞}[\frac{sin\frac{a}{2^n}}{(\frac{a}{2^n})}]

Let, y = (a/2n)

If n→∞, y→0

= (a/2).Limy→0[siny/y]

= (a/2)

Question 26. limn→∞[sin(a/2n)/sin(b/2n)]

Solution:

We have,

limn→∞[sin(a/2n)/sin(b/2n)]

=\lim_{n\to∞}[\frac{sin(\frac{a}{2^n})}{(\frac{a}{2^n})}×(\frac{a}{2^n})]\lim_{n\to∞}[\frac{}{\frac{sin(\frac{b}{2^n})}{(\frac{b}{2^n})}×(\frac{b}{2^n})}]

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

If n→∞, y→0 and z→0

=\frac{y}{z}\lim_{y\to0}[\frac{siny}{y}]\lim_{z\to0}[\frac{y}{siny}]

=\frac{\frac{a}{2^n}}{\frac{b}{2^n}}

= (a/b)

Question 27. limx→-1[(x2 – x – 2)/{(x2 + x) + sin(x + 1)}]

Solution:

We have,

limx→-1[(x2 – x – 2)/{(x2 + x) + sin(x + 1)}]

= limx→-1[(x2 – x – 2)/{x(x + 1) + sin(x + 1)}]

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

Let, y = x + 1

If x→-1, then y→0

= limy→0[y(y – 3)/{y(y – 1) + siny}]

=\lim_{y\to0}[\frac{(y-3)}{(y-1)+\frac{siny}{y}}]

= (0 – 3)/{(0 – 1) + 1}

= -3/0

= ∞

Question 28. limx→2[(x2 – x – 2)/{(x2 – 2x) + sin(x – 2)}]

Solution:

We have,

limx→2[(x2 – x – 2)/{(x2 – 2x) + sin(x – 2)}]

= limx→2[{(x – 2)(x + 1)}/{x(x + 1) + sin(x + 1)}]

Let, y = x – 2

If x→2, then y→0

= limy→0[y(y + 3)/{y(y + 2) + siny}]

\lim_{y\to0}[\frac{(y+3)}{(y+2)+\frac{siny}{y}}]

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

= 3/3

= 1

Question 29. limx→1[(1 – x)tan(πx/2)]

Solution:

We have,

 limx→1[(1 – x)tan(πx/2)]

Here, x→1, h→0

= limh→0[{1 – (1 – h)}tan{π/2(1 – h)}]

= limh→0[htan{π/2-πh/2)}

= limh→0[hcot(πh/2)]

=\lim_{h\to0}[\frac{h}{tan(\frac{πh}{2})}]

=\lim_{h\to0}[\frac{1}{\frac{tan(\frac{πh}{2})}{\frac{πh}{2}}×\frac{π}{2}}]

=\frac{1}{\frac{π}{2}}

= (2/π)

Question 30. limx→π/4[(1 – tanx)/(1 – √2sinx)]

Solution:

We have,

limx→π/4[(1 – tanx)/(1 – √2sinx)]

On rationalizing the denominator.

= limx→π/4[{(1 – tanx)(1 – √2sinx)}/(1 – 2sin2x)]

=\lim_{x\to \frac{π}{4}}[\frac{(1-\frac{sinx}{cosx})(1+\sqrt2sinx)}{cos2x}]

=\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.cos2x}]

=\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.(cos^2x-sin^2x)}]

=\lim_{x\to \frac{π}{4}}[\frac{(cosx-sinx)(1+\sqrt2sinx)}{cosx.(cosx-sinx)(cosx+sinx)}]

=\frac{1+\sqrt2×\frac{1}{\sqrt2}}{(\frac{1}{\sqrt2})(\frac{1}{\sqrt2}+\frac{1}{\sqrt2})}

= 2/1

= 2

Question 31. limx→π[{√(2 + cosx) – 1}/(π – x)2]

Solution:

We have,

limx→π[{√(2 + cosx) – 1}/(π – x)2]

Let, y = [π – x]

Here, x→π, y→0

=\lim_{y\to0}[\frac{\sqrt{2+cos(π-y)}-1}{y^2}]

= limy→0[{√(2 – cosy) – 1}/y2]

On rationalizing the numerator, we get

=\lim_{y\to0}[\frac{2-cosy-1}{y^2(\sqrt{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 × (1/4) × {1/(1 + 1)}

= (1/4)

Question 32. limx→π/4[(√cosx – √sinx)/(x – π/4)]

Solution:

We have,

limx→π/4[(√cosx – √sinx)/(x – π/4)]

On rationalizing the numerator, we get

= limx→π/4[(cosx – sinx)/{(√cosx + √sinx)(x – π/4)}]

\lim_{h\to0}[\frac{cos(\frac{π}{4}+h)-sin(\frac{π}{4}+h)}{(\frac{π}{4}+h-\frac{π}{4})(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)}}]

\lim_{h\to0}[\frac{-2×\frac{1}{\sqrt2}×sinh}{h(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)})}]

\lim_{h\to0}[\frac{-\sqrt2×sinh}{h(\sqrt{cos(\frac{π}{4}+h)}+\sqrt{sin(\frac{π}{4}+h)})}]

=-\sqrt2[\frac{1}{(\frac{1}{\sqrt2})^{\frac{1}{2}}+(\frac{1}{\sqrt2})^{\frac{1}{2}}}]

Question 33. limx→1[(1 – 1/x)/sinπ(x – 1)]

Solution:

We have,

limx→1[(1 – 1/x)/sinπ(x – 1)]

= limx→1[(x – 1)/x{sinπ(x – 1)}]

Let, y = x – 1

If x→1, then y→0

= limy→0[y/{(y + 1)sin(πy)}]

\lim_{y\to0}[\frac{1}{\frac{(y+1).sin(πy)}{y}}]

\lim_{y\to0}[\frac{1}{\frac{(y+1).sin(πy)}{πy}×π}]

= 1/{(1 + 0) × 1 × π}

= 1/π

Question 34. limx→π/6[(cot2x – 3)/(cosecx – 2)]

Solution:

We have,

limx→π/6[(cot2x – 3)/(cosecx – 2)]

= limx→π/6[(cosec2x – 1 – 3)/(cosecx – 2)]

= limx→π/6[(cosec2x – 22)/(cosecx – 2)]

= limx→π/6[{(cosecx + 2)(cosecx – 2)}/(cosecx – 2)]

= limx→π/6[(cosecx + 2)]

= cosec(π/6) + 2

= 2 + 2

= 4

Question 35. limx→π/4[(√2 – cosx – sinx)/(4x – π)2]

Solution:

We have,

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

= limx→π/4[(√2 – cosx – sinx)/{42(π/4 – x)2}]

=\lim_{x\to \frac{π}{2}}\frac{\sqrt2[1-(\frac{1}{\sqrt2}.cosx+\frac{1}{\sqrt2}.sinx)]}{4^2(\frac{π}{4}-x)^2}

=\lim_{x\to \frac{π}{2}}\frac{\sqrt{2}[1-cos(\frac{π}{4}-x)]}{4^2(\frac{π}{4}-x)^2}

=\lim_{x\to \frac{π}{2}}\frac{[2\sqrt{2}sin^2\frac{(\frac{π}{4}-x)}{2}]}{4^2(\frac{π}{4}-x)^2}

=2\sqrt{2}\lim_{x\to \frac{π}{2}}\frac{[sin^2\frac{(\frac{π}{4}-x)}{2}]}{4^2×\frac{(\frac{π}{4}-x)}{4}^2×4}

= 2√2/43

= (2√2 × √2)/(43√2)

= 4/(43√2)

= 1/(16√2)

Question 36. limx→π/2[{(π/2 – x)sinx – 2cosx}/{(π/2 – x) + cotx}]

Solution:

We have,

 limx→π/2[{(π/2 – x)sinx – 2cosx}/{(π/2 – x) + cotx}]

=\lim_{h\to0}\frac{[\frac{π}{2}-(\frac{π}{2}-h)]sin(\frac{π}{2}-h)-2cos(\frac{π}{2}-h)}{[\frac{π}{2}-(\frac{π}{2}-h)]+cot(\frac{π}{2}-h)}

= limh→0[(hcosh-2sinh)/(h+tanh)]

\lim_{h\to0}[\frac{cosh-2\frac{sinh}{h}}{1+\frac{tanh}{h}}]     (On dividing the numerator and denominator by h)

= (1 – 2)/(1 + 1)

= -1/2

Question 37. limx→π/4[(cosx – sinx)/{(π/4 – x)(cosx + sinx)}]

Solution:

We have,

limx→π/4[(cosx – sinx)/{(π/4 – x)(cosx + sinx)}]

On dividing the numerator and denominator by √2, we get

\lim_{x\to \frac{π}{4}}[\frac{\frac{cosx}{\sqrt2}-\frac{sinx}{\sqrt2}}{(\frac{π}{4}-x)(\frac{cosx+sinx}{\sqrt2})}]

\lim_{x\to \frac{π}{4}}[\frac{{\sqrt2}(sin\frac{π}{4}.cosx-cos\frac{π}{4}.sinx)}{(\frac{π}{4}-x)(cosx+sinx)}]

\lim_{x\to \frac{π}{4}}[\frac{{\sqrt2}[sin(\frac{π}{4}-x)]}{(\frac{π}{4}-x)(cosx+sinx)}]

\sqrt2\lim_{x\to \frac{π}{4}}[\frac{[sin(\frac{π}{4}-x)]}{(\frac{π}{4}-x)}]\lim_{x\to \frac{π}{4}}[\frac{1}{(cosx+sinx)}]

\frac{\sqrt2}{\frac{1}{\sqrt2}+\frac{1}{\sqrt2}}

= (√2 × √2)/2

= 1

Question 38. limx→π[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Solution:

We have,

limx→π[{1 – sin(x/2)}/{cos(x/2)(cosx/4 – sinx/4}]

Let, x = π + h

If x→π, then h→0

\lim_{h\to0}[\frac{1-sin(\frac{π+h}{2})}{cos(\frac{π+h}{2})[cos(\frac{π+h}{4})-sin(\frac{π+h}{4})]}]

=\lim_{h\to0}[\frac{1-sin(\frac{π}{2}+\frac{h}{2})}{cos(\frac{π}{4}+\frac{h}{2})[cos(\frac{π}{4}+\frac{h}{4})-sin(\frac{π}{4}+\frac{h}{4})]}]

\lim_{h\to0}[\frac{1-cos(\frac{h}{2})}{-sin(\frac{h}{2})[-\sqrt{2}sin(\frac{h}{2})]}]

\lim_{h\to0}[\frac{2sin^2(\frac{h}{4})}{-sin(\frac{h}{2})[-\sqrt{2}sin(\frac{h}{4})]}]

\lim_{h\to0}[\frac{2sin^2(\frac{h}{4})}{2sin(\frac{h}{4})cos(\frac{h}{4})[\sqrt{2}sin(\frac{h}{4})]}]

(\frac{1}{\sqrt2})\lim_{h\to0}\frac{1}{cos(\frac{h}{4})}

= 1/√2


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