第11类RD Sharma解决方案–第29章限制–练习29.6 |套装2

问题14. Lim _n→∞ {1 ² + 2 ² +……………。 + n ² } /(n ³ )

解决方案：

We have,

Lim_n→∞{1² + 2² + ……………. + n²}/(n³)

= Lim_n→∞[n(n+1)(2n+1)]/6n³

= Lim_n→∞[(n+1)(2n+1)]/6n²

=$\lim_{n\to∞}\frac{(n+1)}{n}\frac{2n+1}{n}\frac{1}{6}$

$\lim_{n\to∞}(1+\frac{1}{n})(2+\frac{1}{n})\frac{1}{6}$

When n → ∞, (1/n) → 0

= 2/6

= 1/3

问题15. Lim _n→∞ {1 + 2 + 3 + 4 +………………。 + n – 1} / n ²

解决方案：

We have,

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

= Lim_n→∞[n(n – 1)/2n²]

= Lim_n→∞[n² – n/2n²]

= Lim_n→∞(1/2 – 1/2n)

When n → ∞, (1/n) → 0

= 1/2

问题16. Lim _n→∞ {1 ³ + 2 ³ +……………。 + n ³ } /(n ⁴ )

解决方案：

We have,

Lim_n→∞{1³ + 2³ + ……………. + n³}/(n⁴)

= Lim_n→∞[n²(n + 1)²]/(4n⁴)           [since (1³ + 2³ + …………. + n³) = n²(n + 1)²/4]

= Lim_n→∞[(n + 1)²]/(4n²)

= Lim_n→∞[(1 + 1/n)² × (1/4)]

When n → ∞, (1/n) → 0

= 1/4

问题17. Lim _n→∞ {1 ³ + 2 ³ +……………。 + n ³ } /(n – 1) ⁴

解决方案：

We have,

Lim_n→∞{1³ + 2³ + ……………. + n³}/(n – 1)⁴

= Lim_n→∞[n²(n + 1)²]/[4(n – 1)⁴]           [since (1³ + 2³ + …………. + n³) = n²(n + 1)²/4]

=$\lim_{n\to∞}\frac{n^4(1+\frac{1}{n})^2}{4n^4(1-\frac{1}{n})^4}$

=$\lim_{n\to∞}\frac{(1+\frac{1}{n})^2}{4(1-\frac{1}{n})^4}$

When n → ∞, (1/n) → 0

= 1/4

问题18. Lim _x→∞ [√x{√(x + 1)–√x}]

解决方案：

We have,

Lim_x→∞[√x{√(x + 1) – √x}]

On rationalizing numerator, we get

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

= Lim_x→∞(√x}/{√(x + 1) + √x}

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

When x → ∞, (1/x) → 0.

= 1/(√1 + 1)

= 1/2

问题19. Lim _x→∞ [1/3 + 1/3 ² + 1/3 ³ +………………+ 1/3 ⁿ ]

解决方案：

We have,

 Lim_x→∞[1/3 + 1/3² + 1/3³ + ……………… + 1/3ⁿ]

This is G.P series of common ratio 1/3.

So, the sum of n temrs of G.P. S_n = [a(1 – rⁿ)]/(1 – r)        (i)

a = 1/3, r = 1/3

On putting the value of a & r in equation (i), we get

S_n = (1/2)(1 – 1/3ⁿ) 

= Lim_x→∞[(1/2)(1 – 1/3ⁿ)]

= (1/2)Lim_x→∞(1 – 1/3ⁿ)

= (1/2)(1 – 0)

= 1/2

问题20. Lim _x→∞ {(x ⁴ + 7x ³ + 46x + a)} / {(x ⁴ + 6)}。

解决方案：

We have,

 Lim_x→∞{(x⁴ + 7x³ + 46x + a)}/{(x⁴ + 6)}

=$\lim_{x\to∞}\frac{x^4(1+\frac{7}{x}+\frac{46}{x^2}+\frac{a}{x^4})}{x^4(1+\frac{6}{x^4})}$

When x → ∞, (1/x), (1/x²), (1/x³), (1/x⁴) → 0

= 1/1

= 1

问题21. f(x)=(ax ² + b)/(x ² +1)，Lim _x→0 f(x)= 1，Lim _x→∞f (x)= 1，然后证明f(- 2)= f(2)= 1

解决方案：

We have,

 f(x) = (ax² + b)/(x² + 1)

= Lim_x→0[(ax² + b)/(x² + 1)]

b/1 = 1

b = 1

= Lim_x→∞[(ax² + b)/(x² + 1)]

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

When x → ∞, (1/x²) → 0.

(a + 0)/(1 + 0) = 1

a = 1

Hence, a = 1, b = 1

f(x) = (x² + 1)/(x² + 1)

f(x) = 1

f(-2) = 1

f(2) = 1   (Since f(x) is independent on x)

f(-2) = f(2) = 1

Hence proved

问题22。证明Lim _x→∞ [√(x ² + x + 1)– x]≠Lim _x→∞ [√(x ² + 1)– x]

解决方案：

We have,

L.H.S,

= Lim_x→∞[√(x² + x + 1) – x]

On rationalizing numerator, we get

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

= Lim_x→∞(x + 1)/[√(x² + x + 1) + x]

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

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

When x → ∞, (1/x), (1/x²) → 0.

= 1/(√1 + 1)

= 1/2

Now we solve R.H.S,

= Lim_x→∞[√(x² + 1) – x]

On rationalizing numerator, we get

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

= Lim_x→∞(1)/[√(x² + 1) + x]

= 1/[√(∞ + 1) + ∞]

= 1/∞

= 0

L.H.S ≠ R.H.S

Hence, Lim_x→∞[√(x² + x + 1) – x] ≠ Lim_x→∞[√(x² + 1) – x]

问题23. Lim _x→-∞ [√(4x ² – 7x)+ 2x]

解决方案：

We have,

 Lim_x→-∞[√(4x² – 7x) + 2x]

Let x = -n when x → -∞, then n → ∞.

= Lim_n→∞[√(4n² + 7n) – 2n]

On rationalizing numerator, we get

= Lim_n→∞[(4n² + 7n) – 4n²]/[√(4n² + 7n) + 2n]

= Lim_n→∞[(7n)/[√(4n² + 7n) + 2n]

= Lim_n→∞(7n)/[n{√(4 + 7/n) + 2}]

= Lim_n→∞(7)/{√(4 + 7/n) + 2}

When n → ∞, (1/n) → 0

= 7/(√4 + 2)

= 7/(2 + 2)

= 7/4

问题24. Lim _x→-∞ [√(x ² – 8x)+ x]

解决方案：

We have,

Lim_x→-∞[√(x² – 8x) + x]

Let x = -n when x → -∞, then n → ∞.

= Lim_n→∞[√(n² + 8n) – n]

On rationalizing numerator, we get

= Lim_n→∞[(n² + 8n) – n²]/[√(n² + 8n) + n]

= Lim_n→∞[(8n)/[√(n² + 8n) + n]

= Lim_n→∞(8n)/[n{√(1 + 8/n) + 1}]

= Lim_n→∞(8)/{√(1 + 8/n) + 1}

When n → ∞, (1/n) → 0

= 8/(√1 + 1)

= 8/2

= 4

问题25. Lim _n→∞ (1 ⁴ + 2 ⁴ +………。+ n ⁴ )/ n ⁵ – Lim _n→∞ (1 ³ + 2 ³ +………。+ n ³ )/ n ⁵

解决方案：

We have,

Lim_n→∞(1⁴ + 2⁴ + ……….+ n⁴)/n⁵ – Lim_n→∞(1³ + 2³ + ………. + n³)/n⁵

=$\lim_{n\to∞}\frac{\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}}{n^5}-\lim_{n\to∞}\frac{\frac{n^2(n+1)^2}{4}}{n^5}$

=$\lim_{n\to∞}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}-\lim_{n\to∞}\frac{n^2(n+1)^2}{4n^5}$

=$\lim_{n\to∞}\frac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}-\lim_{n\to∞}\frac{(n+1)^2}{4n^3}$

=$\frac{1}{30}\lim_{n\to∞}(1+\frac{1}{n})(2+\frac{1}{n})(3+\frac{3}{n}-\frac{1}{n^2})-\frac{1}{4}\lim_{n\to∞}\frac{1}{n}(1+\frac{1}{n})^2$

When n → ∞, (1/n), (1/n²), (1/n³) → 0

= 1/3 × 1 × 2 × 3 – 1/4 × 0

= 6/30

= 1/5

问题26. Lim _n→∞ {(1.2 + 2.3 + 3.4 +………。+ n(n + 1)} / n ³

解决方案：

We have,

Lim_n→∞{(1.2 + 2.3 + 3.4 + ……….+ n (n + 1)}/n³

=$\lim_{n\to∞}\frac{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}}{n^3}$

=$\lim_{n\to∞}\frac{n(n+1)[\frac{(2n+1)+3}{6}]}{n^3}$

=$\lim_{n\to∞}\frac{n(n+1)[\frac{(2n+4)}{6}]}{n^3}$

=$\lim_{n\to∞}\frac{n(n+1)[(2n+4)]}{6n^3}$

=$\lim_{n\to∞}\frac{n(n+1)[(2n+4)]}{6n^3}$

=$\lim_{n\to∞}\frac{(1+\frac{1}{n})(2+\frac{4}{n})}{6}$

When n → ∞, (1/n) → 0

= (1 × 2)/6

= 2/6

= 1/3