Class 11 RD Sharma解决方案–第29章限制–练习29.8 |套装2

问题20. lim _x→1 [(1 +cosπx)/(1-x) ² ]

解决方案：

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

问题21. lim _x→1 [(1-x ² )/sinπx]

解决方案：

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

问题22. lim _{x→π/ 4} [(1-sin2x)/(1 + cos4x)]

解决方案：

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)

问题23. lim _x→π [(1 + cosx)/ tan ² x]

解决方案：

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

问题24. lim _n→∞ [nsin(π/ 4n)cos(π/ 4n)]

解决方案：

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)

问题25. lim _n→∞ [2 ^n-1 sin(a / 2 ⁿ )]

解决方案：

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)

问题26. lim _n→∞ [sin(a / 2 ⁿ )/ sin(b / 2 ⁿ )]

解决方案：

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)

问题27. lim _x→-1 [(x ² – x – 2)/ {(x ² + x)+ sin(x + 1)}]

解决方案：

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

= ∞

问题28. lim _x→2 [(x ² – x – 2)/ {(x ² – 2x)+ sin(x – 2)}]

解决方案：

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

问题29. lim _x→1 [(1-x)tan(πx/ 2)]

解决方案：

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)

问题30. lim _{x→π/ 4} [(1-tanx)/(1-√2sinx)]

解决方案：

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

问题31. lim _x→π [{√(2 + cosx)– 1} /(π– x) ² ]

解决方案：

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)

问题32. lim _{x→π/ 4} [(√cosx–√sinx)/(x –π/ 4)]

解决方案：

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)}]

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

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

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

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

问题33. lim _x→1 [(1 – 1 / x)/sinπ(x – 1)]

解决方案：

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

问题34. lim _{x→π/ 6} [(cot ² x – 3)/(cosecx – 2)]

解决方案：

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

问题35. lim _{x→π/ 4} [(√2-cosx-sinx)/(4x-π) ² ]

解决方案：

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)

问题36. lim _{x→π/ 2} [{(π/ 2 – x)sinx – 2cosx} / {(π/ 2 – x)+ cotx}]

解决方案：

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

问题37. lim _{x→π/ 4} [(cosx-sinx)/ {(π/ 4-x)(cosx + sinx)}]

解决方案：

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

问题38. lim _x→π [{1-sin(x / 2)} / {cos(x / 2)(cosx / 4-sinx / 4}]

解决方案：

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