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

Question 1. Lim_x→∞{(3x - 1)(4x - 2)}/{(x + 8)(x - 1)}.

Solution:

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

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

Solution:

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

Question 3. Lim_x→∞{(5x³ - 6)}/{√(9 + 4x⁶)}.

Solution:

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

Question 4. Lim_x→∞{√(x² + cx) - x}

Solution:

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

Question 5. Lim_x→∞{√(x + 1) - √x}

Solution:

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

Question 6. Lim_x→∞{√(x² + 7x) - x}

Solution:

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

Question 7.  Lim_x→∞(x)/{√(4x² + 1) - 1}

Solution:

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

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

Solution:

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

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

Solution:

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

Question 10. Lim_x→∞{√(x² + a²) - √(x² + b²)}/{√(x² + c²) - √(x² + d²)}

Solution:

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)

Question 11. Lim_n→∞{(n + 2)! + (n + 1)!}/{(n + 2)! - (n + 1)!}.

Solution:

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

Question 12. Lim_x→∞[x{√(x² + 1) - √(x² - 1)}]

Solution:

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

Question 13.  Lim_x→∞[√(x + 2){√(x + 1) - √x}]

Solution:

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