Class 11 RD Sharma解决方案–第29章限制–练习29.8 |套装1

问题1. lim _{x→π/ 2} [π/ 2 – x] .tanx

解决方案：

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

问题2. lim _{x→π/ 2} [sin2x / cosx]

解决方案：

We have,

lim_x→π/2[sin2x/cosx]

= lim_x→π/2[2sinx.cosx/cosx]

= 2Lim_x→π/2[sinx]

= 2

问题3. lim _{x→π/ 2} [cos ² x /(1 – sinx)]

解决方案：

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

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

解决方案：

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

问题5. lim _x→a [(cosx-cosa)/(x- a)]

解决方案：

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

问题6. lim _{x→π/ 4} [(1-tanx)/(x-π/ 4)]

解决方案：

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

问题7. lim _{x→π/ 2} [(1-sinx)/(π/ 4-x) ² ]

解决方案：

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)

问题8. lim _{x→π/ 3} [(√3-tanx)/(π-3x)]

解决方案：

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)

问题9. lim _x→a [(asinx – xsina)/(ax ² – xa ² )]

解决方案：

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]

问题10. lim _{x→π/ 2} [{√2–√(1 + sinx)} / cos ² x]

解决方案：

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

问题11。lim _{x→π/ 2} [{√(2 – sinx)– 1} /(π/ 2 – x) ² ]

解决方案：

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

问题12. lim _{x→π/ 4} [(√2-cosx-sinx)/(π/ 4-x) ² ]

解决方案：

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)

问题13. lim _{x→π/ 8} [(cot4x – cos4x)/(π– 8x) ³ ]

解决方案：

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

问题14. lim _x→a [(cosx – cosa)/(√x–√a)]

解决方案：

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

问题_{15。lim x→π} [{√(5 + cosx)– 2} /(π– x) ² ]

解决方案：

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)

问题_{16。lim x→a} [(cos√x–cos√a)/(x – a)]

解决方案：

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)

问题17. lim _x→a [(sin√x–sin√a)/(x – a)]

解决方案：

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)

问题18. lim _x→1 [(1-x ² )/sin2πx]

解决方案：

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

问题19. lim _{x→π/ 4} [{f(x)– f(π/ 4)} / {x –π/ 4}]

解决方案：

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