第11类RD Sharma解决方案–第29章限制–练习29.6 |套装1

问题1. Lim _x→∞ {(3x – 1)(4x – 2)} / {(x + 8)(x – 1)}。

解决方案：

We have,

Lim_x→∞{(3x – 1)(4x – 2)}/{(x + 8)(x – 1)}

= $\lim_{x\to ∞}\frac{x(3-\frac{1}{x})x(4-\frac{2}{x})}{x(1+\frac{8}{x})x(1-\frac{1}{x})}$

= $\lim_{x\to∞}\frac{(3-\frac{1}{x})(4-\frac{2}{x})}{(1+\frac{8}{x})(1-\frac{1}{x})}$

When x → ∞, (1/x) → 0.

= (3 × 4)/(1 × 1)

= 12

问题2. Lim _x→∞ {(3x ³ – 4x ² + 6x – 1)} / {(2x ³ + x ² – 5x + 7)}。

解决方案：

We have,

Lim_x→∞{(3x³ – 4x² + 6x – 1)}/{(2x³ + x² – 5x + 7)}

=$\lim_{x\to∞}\frac{x^3(3-\frac{4}{x}+\frac{6}{x^2}-\frac{1}{x^3})}{x^3(2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3})}$

=$\lim_{x\to∞}\frac{(3-\frac{4}{x}+\frac{6}{x^2}-\frac{1}{x^3})}{(2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3})}$

When x → ∞, (1/x), (1/x²), (1/x³) → 0.

= 3/2

问题3. Lim _x→∞ {(5x ³ – 6)} / {√(9 + 4x ⁶ )}。

解决方案：

We have,

Lim_x→∞{(5x³ – 6)}/{√(9 + 4x⁶)}

=$\lim_{x\to∞}\frac{x^3(5-\frac{6}{x^3})}{x^3\sqrt{(4+\frac{9}{x^6}})}$

= $\lim_{x\to∞}\frac{(5-\frac{6}{x^3})}{\sqrt{(4+\frac{9}{x^6}})}$

When x → ∞, (1/x), (1/x³) → 0.

= 5/√4

= 5/2

问题4. Lim _x→∞ {√(x ² + cx)– x}

解决方案：

We have,

Lim_x→∞{√(x²+cx)-x}

On rationalizing numerator, we get

= Lim_x→∞{(x² + cx) – x²}/{√(x² + cx) + x}

= Lim_x→∞(cx)/{√(x² + cx) + x}

= Lim_x→∞(cx)/[x{√(x + c/x) + 1}]

= Lim_x→∞(c)/{√(1 + c/x) + 1}

When x → ∞, (1/x) → 0.

= c/(√1 + 1)

= c/2

问题5. Lim _x→∞ {√(x + 1)–√x}

解决方案：

We have,

Lim_x→∞{√(x + 1) – √x}

On rationalizing numerator, we get

= Lim_x→∞{(x+1)-x}/{√(x+1)+√x}

= Lim_x→∞(1)/{√(x+1)+√x}

= $\lim_{x\to∞}\frac{1}{\sqrt{x}(1+\frac{1}{x}+1)}$

When x → ∞, (1/x) → 0.

= 0

问题6. Lim _x→∞ {√(x ² + 7x)– x}

解决方案：

We have,

Lim_x→∞{√(x² + 7x) – x}

On rationalizing numerator, we get

= Lim_x→∞{(x²+7x)-x2}/{√(x²+7x)+x}

= Limx→∞(7x)/{√(x²+7x)+x}

=$\lim_{x\to∞}\frac{7x}{x[\sqrt{(1+\frac{7}{x}})+1]}$

=$\lim_{x\to∞}\frac{7}{[\sqrt{(1+\frac{7}{x}})+1]}$

When x → ∞, (1/x) → 0.

= 7/(√1 + 1)

= 7/2

问题7. Lim _x→∞ (x)/ {√(4x ² +1)– 1}

解决方案：

We have,

Lim_x→∞(x)/{√(4x² + 1) – 1}

Rationalising denominator.

= Lim_x→∞[x{√(4x² + 1) + 1}]/{(4x² + 1) – 1}

= Lim_x→∞[x{√(4x² + 1) + 1}]/(4x²)

= Lim_x→∞[{√(4x² + 1) + 1}]/(4x)

=$\lim_{x\to∞}\frac{\sqrt{4+\frac{1}{x^2}}}{4}$

When x → ∞, (1/x²) → 0.

= √4/4

= 2/4

= 1/2

问题8. Lim _n→∞ (n ² )/ {1 + 2 + 3 + 4 +……………。 + n}

解决方案：

We have,

Lim_n→∞(n²)/{1 + 2 + 3 + 4 + ……………. + n}

=$\lim_{n\to∞}\frac{n^2}{\frac{n(n+1)}{2}}$

= Lim_n→∞(2n)/(n+1)

= Lim_n→∞(2)/(1+1/n)

When n → ∞, (1/n) → 0

= 2/(1 + 0)

= 2

问题9. Lim _x→∞ (3x ^-1 + 4x ^-2 )/(5x ^-1 + 6x ^-2 )

解决方案：

We have,

Lim_x→∞(3x^-1 + 4x^-2)/(5x^-1 + 6x^-2)

= $\lim_{x\to∞}\frac{\frac{3}{x}+\frac{4}{x^2}}{\frac{5}{x}+\frac{6}{x^2}}$

= $\lim_{x\to∞}\frac{\frac{1}{x}(3+\frac{4}{x})}{\frac{1}{x}(5+\frac{6}{x})}$

When x → ∞, (1/x) → 0.

= 3/5

问题10. Lim _x→∞ {√(x ² + a ² )–√(x ² + b ² )} / {√(x ² + c ² )–√(x ² + d ² )}

解决方案：

We have,

 Lim_x→∞{√(x² + a²) – √(x² + b²)}/{√(x² + c²) – √(x² + d²)}

On rationalizing numerator and denominator, we get

=$\lim_{x\to∞}\frac{(\sqrt{x^2+a^2}-\sqrt{x^2+b^2})(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})(\sqrt{x^2+c^2}+\sqrt{x^2+d^2})}{(\sqrt{x^2+c^2}-\sqrt{x^2+d^2})(\sqrt{x^2+c^2}+\sqrt{x^2+d^2})(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$

=$\lim_{x\to∞}\frac{(x^2+a^2)-(x^2+b^2))(\sqrt{x^2+c^2}+\sqrt{x^2+d^2})}{(x^2+c^2)-(x^2+d^2)(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$

=$=\lim_{x\to∞}\frac{(a^2-b^2)(\sqrt{x^2+c^2}+\sqrt{x^2+d^2})}{(c^2-d^2)(\sqrt{x^2+a^2}+\sqrt{x^2+b^2})}$

=$\lim_{x\to∞}\frac{(a^2-b^2)\frac{1}{x}(\sqrt{(1+\frac{c^2}{x^2}})+\sqrt{(1+\frac{d^2}{x^2}}}{(c^2-d^2)\frac{1}{x}(\sqrt{1+\frac{a^2}{x^2}}+\sqrt{1+\frac{b^2}{x^2}})}$

When x → ∞, (1/x²) → 0.

=$\frac{a^2-b^2}{c^2-d^2}×\frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}$

= (a² – b²)/(c² – d²)

问题11：Lim _n→∞ {(n + 2)！ +(n + 1)！} / {(n + 2)！ –(n + 1)！}。

解决方案：

We have,

Lim_n→∞{(n + 2)! + (n + 1)!}/{(n + 2)! – (n + 1)!}

= Lim_n→∞{(n + 2)(n + 1)! + (n + 1)!}/{(n + 2)(n + 1)! – (n + 1)!}

= Lim_n→∞[(n + 1)!{(n + 2) + 1}]/[(n + 1)!{(n + 2) – 1}]

= Lim_n→∞(n + 3)/(n + 1)

= Lim_n→∞[n(1 + 3/n)]/[n(1 + 1/n)]

When n → ∞, (1/n) → 0.

= 1/1

= 1

问题12. Lim _x→∞ [x {√(x ² +1)–√(x ² – 1)}]

解决方案：

We have,

Lim_x→∞[x{√(x² + 1) – √(x² – 1)}]

On rationalizing numerator, we get

= Lim_x→∞[x{(x² + 1) – (x² – 1)}]/{√(x² + 1) + √(x² – 1)}

= Lim_x→∞(2x)/{√(x² + 1) + √(x² – 1)}

= Lim_x→∞(2x)/[x{√(1 + 1/x²) + √(1 – 1/x²)}]

= Lim_x→∞(2)/[{√(1 + 1/x²) + √(1 – 1/x²)}]

When x → ∞, (1/x²) → 0.

= 2/(√1 + √1)

= 2/2

= 1

问题13. Lim _x→∞ [√(x + 2){√(x + 1)–√x}]

解决方案：

We have,

 Lim_x→∞[√(x + 2){√(x + 1) – √x}]

On rationalizing numerator, we get

= Lim_x→∞[√(x + 2){(x + 1) – x}]/{√(x + 1) + √x}

= Lim_x→∞[√(x + 2)]/{√(x + 1) + √x}

= Lim_x→∞[x√(1 + 2/x)]/[x{√(1 + 1/x) + √1}]

= Lim_x→∞[√(1 + 2/x)]/{√(1 + 1/x) + √1}

When x → ∞, (1/x) → 0.

= 1/(√1 + √1)

= 1/2