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

Last Updated : 18 Apr, 2022

### Question 14.  Limnâ†’âˆž{12 + 22 + ……………. + n2}/(n3)

Solution:

We have,

Limnâ†’âˆž{12 + 22 + ……………. + n2}/(n3)

= Limnâ†’âˆž[n(n+1)(2n+1)]/6n3

= Limnâ†’âˆž[(n+1)(2n+1)]/6n2

=

When n â†’ âˆž, (1/n) â†’ 0

= 2/6

= 1/3

### Question 15. Limnâ†’âˆž{1 + 2 + 3 + 4 +……………. + n – 1}/n2

Solution:

We have,

Limnâ†’âˆž{1 + 2 + 3 + 4 +……………. + n – 1}/n2

= Limnâ†’âˆž[n(n – 1)/2n2]

= Limnâ†’âˆž[n2 – n/2n2]

= Limnâ†’âˆž(1/2 – 1/2n)

When n â†’ âˆž, (1/n) â†’ 0

= 1/2

### Question 16. Limnâ†’âˆž{13 + 23 + ……………. + n3}/(n4)

Solution:

We have,

Limnâ†’âˆž{13 + 23 + ……………. + n3}/(n4)

= Limnâ†’âˆž[n2(n + 1)2]/(4n4)           [since (13 + 23 + …………. + n3) = n2(n + 1)2/4]

= Limnâ†’âˆž[(n + 1)2]/(4n2)

= Limnâ†’âˆž[(1 + 1/n)2 Ã— (1/4)]

When n â†’ âˆž, (1/n) â†’ 0

= 1/4

### Question 17.  Limnâ†’âˆž{13 + 23 + ……………. + n3}/(n – 1)4

Solution:

We have,

Limnâ†’âˆž{13 + 23 + ……………. + n3}/(n – 1)4

= Limnâ†’âˆž[n2(n + 1)2]/[4(n – 1)4]           [since (13 + 23 + …………. + n3) = n2(n + 1)2/4]

=

=

When n â†’ âˆž, (1/n) â†’ 0

= 1/4

### Question 18. Limxâ†’âˆž[âˆšx{âˆš(x + 1) – âˆšx}]

Solution:

We have,

Limxâ†’âˆž[âˆšx{âˆš(x + 1) – âˆšx}]

On rationalizing numerator, we get

= Limxâ†’âˆž[(âˆšx){(x + 1) – x}]/{âˆš(x + 1) + âˆšx}

= Limxâ†’âˆž(âˆšx}/{âˆš(x + 1) + âˆšx}

=

When x â†’ âˆž, (1/x) â†’ 0.

= 1/(âˆš1 + 1)

= 1/2

### Question 19. Limxâ†’âˆž[1/3 + 1/32 + 1/33 + ……………… + 1/3n]

Solution:

We have,

Limxâ†’âˆž[1/3 + 1/32 + 1/33 + ……………… + 1/3n]

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

So, the sum of n terms 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

= Limxâ†’âˆž[(1/2)(1 – 1/3n)]

= (1/2)Limxâ†’âˆž(1 – 1/3n)

= (1/2)(1 – 0)

= 1/2

### Question 20. Limxâ†’âˆž{(x4 + 7x3 + 46x + a)}/{(x4 + 6)}.

Solution:

We have,

Limxâ†’âˆž{(x4 + 7x3 + 46x + a)}/{(x4 + 6)}

=

When x â†’ âˆž, (1/x), (1/x2), (1/x3), (1/x4) â†’ 0

= 1/1

= 1

### Question 21. f(x) = (ax2 + b)/(x2 + 1), Limxâ†’0f(x) = 1, Limxâ†’âˆžf(x) = 1, then prove that f(-2) = f(2) = 1

Solution:

We have,

f(x) = (ax2 + b)/(x2 + 1)

= Limxâ†’0[(ax2 + b)/(x2 + 1)]

b/1 = 1

b = 1

= Limxâ†’âˆž[(ax2 + b)/(x2 + 1)]

When x â†’ âˆž, (1/x2) â†’ 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 Limxâ†’âˆž[âˆš(x2 + x + 1) – x] â‰  Limxâ†’âˆž[âˆš(x2 + 1) – x]

Solution:

We have,

L.H.S,

= Limxâ†’âˆž[âˆš(x2 + x + 1) – x]

On rationalizing numerator, we get

= Limxâ†’âˆž[(x2 + x + 1) – x2]/[âˆš(x2 + x + 1) + x]

= Limxâ†’âˆž(x + 1)/[âˆš(x2 + x + 1) + x]

= Limxâ†’âˆž[x(1 + 1/x)/[x{âˆš(1 + 1/x + 1/x2) + 1}]

= Limxâ†’âˆž[(1 + 1/x)/[{âˆš(1 + 1/x + 1/x2) + 1}]

When x â†’ âˆž, (1/x), (1/x2) â†’ 0.

= 1/(âˆš1 + 1)

= 1/2

Now we solve R.H.S,

= Limxâ†’âˆž[âˆš(x2 + 1) – x]

On rationalizing numerator, we get

= Limxâ†’âˆž[(x2 + 1) – x2]/[âˆš(x2 + 1) + x]

= Limxâ†’âˆž(1)/[âˆš(x2 + 1) + x]

= 1/[âˆš(âˆž + 1) + âˆž]

= 1/âˆž

= 0

L.H.S â‰  R.H.S

Hence, Limxâ†’âˆž[âˆš(x2 + x + 1) – x] â‰  Limxâ†’âˆž[âˆš(x2 + 1) – x]

### Question 23. Limxâ†’-âˆž[âˆš(4x2 – 7x) + 2x]

Solution:

We have,

Limxâ†’-âˆž[âˆš(4x2 – 7x) + 2x]

Let x = -n when x â†’ -âˆž, then n â†’ âˆž.

= Limnâ†’âˆž[âˆš(4n2 + 7n) – 2n]

On rationalizing numerator, we get

= Limnâ†’âˆž[(4n2 + 7n) – 4n2]/[âˆš(4n2 + 7n) + 2n]

= Limnâ†’âˆž[(7n)/[âˆš(4n2 + 7n) + 2n]

= Limnâ†’âˆž(7n)/[n{âˆš(4 + 7/n) + 2}]

= Limnâ†’âˆž(7)/{âˆš(4 + 7/n) + 2}

When n â†’ âˆž, (1/n) â†’ 0

= 7/(âˆš4 + 2)

= 7/(2 + 2)

= 7/4

### Question 24. Limxâ†’-âˆž[âˆš(x2 – 8x) + x]

Solution:

We have,

Limxâ†’-âˆž[âˆš(x2 – 8x) + x]

Let x = -n when x â†’ -âˆž, then n â†’ âˆž.

= Limnâ†’âˆž[âˆš(n2 + 8n) – n]

On rationalizing numerator, we get

= Limnâ†’âˆž[(n2 + 8n) – n2]/[âˆš(n2 + 8n) + n]

= Limnâ†’âˆž[(8n)/[âˆš(n2 + 8n) + n]

= Limnâ†’âˆž(8n)/[n{âˆš(1 + 8/n) + 1}]

= Limnâ†’âˆž(8)/{âˆš(1 + 8/n) + 1}

When n â†’ âˆž, (1/n) â†’ 0

= 8/(âˆš1 + 1)

= 8/2

= 4

### Question 25. Limnâ†’âˆž(14 + 24 + ……….+ n4)/n5 – Limnâ†’âˆž(13 + 23 + ………. + n3)/n5

Solution:

We have,

Limnâ†’âˆž(14 + 24 + ……….+ n4)/n5 – Limnâ†’âˆž(13 + 23 + ………. + n3)/n5

=

=

=

=

When n â†’ âˆž, (1/n), (1/n2), (1/n3) â†’ 0

= 1/3 Ã— 1 Ã— 2 Ã— 3 – 1/4 Ã— 0

= 6/30

= 1/5

### Question 26. Limnâ†’âˆž{(1.2 + 2.3 + 3.4 + ……….+ n (n + 1)}/n3

Solution:

We have,

Limnâ†’âˆž{(1.2 + 2.3 + 3.4 + ……….+ n (n + 1)}/n3

=

=

=

=

=

=

When n â†’ âˆž, (1/n) â†’ 0

= (1 Ã— 2)/6

= 2/6

= 1/3

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