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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.6 | Set 2

  • Difficulty Level : Easy
  • Last Updated : 30 Apr, 2021
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Question 14.  Limn→∞{12 + 22 + ……………. + n2}/(n3)

Solution:

We have,

Limn→∞{12 + 22 + ……………. + n2}/(n3)

= Limn→∞[n(n+1)(2n+1)]/6n3

= Limn→∞[(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

Question 15. Limn→∞{1 + 2 + 3 + 4 +……………. + n – 1}/n2

Solution:

We have,

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



= Limn→∞[n(n – 1)/2n2]

= Limn→∞[n2 – n/2n2]

= Limn→∞(1/2 – 1/2n)

When n → ∞, (1/n) → 0

= 1/2

Question 16. Limn→∞{13 + 23 + ……………. + n3}/(n4)

Solution:

We have,

Limn→∞{13 + 23 + ……………. + n3}/(n4)

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

= Limn→∞[(n + 1)2]/(4n2)



= Limn→∞[(1 + 1/n)2 × (1/4)]

When n → ∞, (1/n) → 0

= 1/4

Question 17.  Limn→∞{13 + 23 + ……………. + n3}/(n – 1)4

Solution:

We have,

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

= Limn→∞[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

Question 18. Limx→∞[√x{√(x + 1) – √x}]

Solution:

We have,

Limx→∞[√x{√(x + 1) – √x}]

On rationalizing numerator, we get

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

= Limx→∞(√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

Question 19. Limx→∞[1/3 + 1/32 + 1/33 + ……………… + 1/3n]

Solution:

We have,

 Limx→∞[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. Sn = [a(1 – rn)]/(1 – r)        (i)

a = 1/3, r = 1/3

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

Sn = (1/2)(1 – 1/3n

= Limx→∞[(1/2)(1 – 1/3n)]



= (1/2)Limx→∞(1 – 1/3n)

= (1/2)(1 – 0)

= 1/2

Question 20. Limx→∞{(x4 + 7x3 + 46x + a)}/{(x4 + 6)}.

Solution:

We have,

 Limx→∞{(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

Question 21. f(x) = (ax2 + b)/(x2 + 1), Limx→0f(x) = 1, Limx→∞f(x) = 1, then prove that f(-2) = f(2) = 1

Solution:

We have,

 f(x) = (ax2 + b)/(x2 + 1)

= Limx→0[(ax2 + b)/(x2 + 1)]

b/1 = 1

b = 1

= Limx→∞[(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

Question 22. Show that Limx→∞[√(x2 + x + 1) – x] ≠ Limx→∞[√(x2 + 1) – x]

Solution:

We have,

L.H.S,

= Limx→∞[√(x2 + x + 1) – x]

On rationalizing numerator, we get

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

= Limx→∞(x + 1)/[√(x2 + x + 1) + x]

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

= Limx→∞[(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,

= Limx→∞[√(x2 + 1) – x]

On rationalizing numerator, we get

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

= Limx→∞(1)/[√(x2 + 1) + x]

= 1/[√(∞ + 1) + ∞]

= 1/∞

= 0

L.H.S ≠ R.H.S

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

Question 23. Limx→-∞[√(4x2 – 7x) + 2x]

Solution:

We have,

 Limx→-∞[√(4x2 – 7x) + 2x]

Let x = -n when x → -∞, then n → ∞.

= Limn→∞[√(4n2 + 7n) – 2n]

On rationalizing numerator, we get

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

= Limn→∞[(7n)/[√(4n2 + 7n) + 2n]

= Limn→∞(7n)/[n{√(4 + 7/n) + 2}]

= Limn→∞(7)/{√(4 + 7/n) + 2}

When n → ∞, (1/n) → 0

= 7/(√4 + 2)

= 7/(2 + 2)

= 7/4

Question 24. Limx→-∞[√(x2 – 8x) + x]

Solution:

We have,

Limx→-∞[√(x2 – 8x) + x]

Let x = -n when x → -∞, then n → ∞.

= Limn→∞[√(n2 + 8n) – n]

On rationalizing numerator, we get

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

= Limn→∞[(8n)/[√(n2 + 8n) + n]

= Limn→∞(8n)/[n{√(1 + 8/n) + 1}]

= Limn→∞(8)/{√(1 + 8/n) + 1}

When n → ∞, (1/n) → 0

= 8/(√1 + 1)

= 8/2

= 4

Question 25. Limn→∞(14 + 24 + ……….+ n4)/n5 – Limn→∞(13 + 23 + ………. + n3)/n5

Solution:

We have,



Limn→∞(14 + 24 + ……….+ n4)/n5 – Limn→∞(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

Question 26. Limn→∞{(1.2 + 2.3 + 3.4 + ……….+ n (n + 1)}/n3

Solution:



We have,

Limn→∞{(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

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