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,

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[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. lim_x→1[(1 - x²)/sinπx]

Solution:

We have,

lim_x→1[(1 - x²)/sinπx]

Here, x→1, h→0

= lim_h→0[{1 - (1 - h)²}/sinπ(1 - h)]

= lim_h→0[(2h - h²)/-sinπh]

= -lim_h→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. 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[2sin²h/2sin²2h]

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

= (1/4)

Question 23. lim_x→π[(1 + cosx)/tan²x]

Solution:

We have,

lim_x→π[(1 + cosx)/tan²x]

Here, x→π, h→0

= lim_h→0[{1+cos(π + h)}/tan²(π + h)]

= lim_h→0[(1 - cosh)/tan²h]

= lim_h→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. 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)]

=\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).Lim_y→0[siny/y]

= (π/4)

Question 25. lim_n→∞[2^n-1sin(a/2ⁿ)]

Solution:

We have,

lim_n→∞[2^n-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).Lim_y→0[siny/y]

= (a/2)

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

Solution:

We have,

lim_n→∞[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. lim_x→-1[(x² - x - 2)/{(x² + x) + sin(x + 1)}]

Solution:

We have,

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

= lim_x→-1[(x² - 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}]

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

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

= -3/0

= ∞

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

Solution:

We have,

lim_x→2[(x² - x - 2)/{(x² - 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}]

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

= (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)]

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

= lim_y→0[{√(2 - cosy) - 1}/y2]

On rationalizing the numerator, we get

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

= \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. 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)}]

= \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. lim_x→π/6[(cot²x - 3)/(cosecx - 2)]

Solution:

We have,

lim_x→π/6[(cot²x - 3)/(cosecx - 2)]

= lim_x→π/6[(cosec²x - 1 - 3)/(cosecx - 2)]

= lim_x→π/6[(cosec²x - 2²)/(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)/{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. 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\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)}

= lim_h→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. 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

= \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. 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

= \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