# Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.6 | Set 1

• Last Updated : 30 Apr, 2021

### Question 1. Limx→∞{(3x – 1)(4x – 2)}/{(x + 8)(x – 1)}.

Solution:

We have,

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

When x → ∞, (1/x) → 0.

= (3 × 4)/(1 × 1)

= 12

### Question 2. Limx→∞{(3x3 – 4x2 + 6x – 1)}/{(2x3 + x2 – 5x + 7)}.

Solution:

We have,

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

=

=

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

= 3/2

### Question 3. Limx→∞{(5x3 – 6)}/{√(9 + 4x6)}.

Solution:

We have,

Limx→∞{(5x3 – 6)}/{√(9 + 4x6)}

=

When x → ∞, (1/x), (1/x3) → 0.

= 5/√4

= 5/2

### Question 4. Limx→∞{√(x2 + cx) – x}

Solution:

We have,

Limx→∞{√(x2+cx)-x}

On rationalizing numerator, we get

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

= Limx→∞(cx)/{√(x2 + cx) + x}

= Limx→∞(cx)/[x{√(x + c/x) + 1}]

= Limx→∞(c)/{√(1 + c/x) + 1}

When x → ∞, (1/x) → 0.

= c/(√1 + 1)

= c/2

### Question 5. Limx→∞{√(x + 1) – √x}

Solution:

We have,

Limx→∞{√(x + 1) – √x}

On rationalizing numerator, we get

= Limx→∞{(x+1)-x}/{√(x+1)+√x}

= Limx→∞(1)/{√(x+1)+√x}

When x → ∞, (1/x) → 0.

= 0

### Question 6. Limx→∞{√(x2 + 7x) – x}

Solution:

We have,

Limx→∞{√(x2 + 7x) – x}

On rationalizing numerator, we get

= Limx→∞{(x2+7x)-x2}/{√(x2+7x)+x}

= Limx→∞(7x)/{√(x2+7x)+x}

=

=

When x → ∞, (1/x) → 0.

= 7/(√1 + 1)

= 7/2

### Question 7.  Limx→∞(x)/{√(4x2 + 1) – 1}

Solution:

We have,

Limx→∞(x)/{√(4x2 + 1) – 1}

Rationalising denominator.

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

= Limx→∞[x{√(4x2 + 1) + 1}]/(4x2)

= Limx→∞[{√(4x2 + 1) + 1}]/(4x)

=

When x → ∞, (1/x2) → 0.

= √4/4

= 2/4

= 1/2

### Question 8. Limn→∞(n2)/{1 + 2 + 3 + 4 + ……………. + n}

Solution:

We have,

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

=

= Limn→∞(2n)/(n+1)

= Limn→∞(2)/(1+1/n)

When n → ∞, (1/n) → 0

= 2/(1 + 0)

= 2

### Question 9. Limx→∞(3x-1 + 4x-2)/(5x-1 + 6x-2)

Solution:

We have,

Limx→∞(3x-1 + 4x-2)/(5x-1 + 6x-2)

When x → ∞, (1/x) → 0.

= 3/5

### Question 10. Limx→∞{√(x2 + a2) – √(x2 + b2)}/{√(x2 + c2) – √(x2 + d2)}

Solution:

We have,

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

On rationalizing numerator and denominator, we get

=

=

=

=

When x → ∞, (1/x2) → 0.

=

= (a2 – b2)/(c2 – d2)

### Question 11. Limn→∞{(n + 2)! + (n + 1)!}/{(n + 2)! – (n + 1)!}.

Solution:

We have,

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

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

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

= Limn→∞(n + 3)/(n + 1)

= Limn→∞[n(1 + 3/n)]/[n(1 + 1/n)]

When n → ∞, (1/n) → 0.

= 1/1

= 1

### Question 12. Limx→∞[x{√(x2 + 1) – √(x2 – 1)}]

Solution:

We have,

Limx→∞[x{√(x2 + 1) – √(x2 – 1)}]

On rationalizing numerator, we get

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

= Limx→∞(2x)/{√(x2 + 1) + √(x2 – 1)}

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

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

When x → ∞, (1/x2) → 0.

= 2/(√1 + √1)

= 2/2

= 1

### Question 13.  Limx→∞[√(x + 2){√(x + 1) – √x}]

Solution:

We have,

Limx→∞[√(x + 2){√(x + 1) – √x}]

On rationalizing numerator, we get

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

= Limx→∞[√(x + 2)]/{√(x + 1) + √x}

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

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

When x → ∞, (1/x) → 0.

= 1/(√1 + √1)

= 1/2

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