Question 1. Limx→∞{(3x - 1)(4x - 2)}/{(x + 8)(x - 1)}.
Solution:
We have,
Limx→∞{(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. Limx→∞{(3x3 - 4x2 + 6x - 1)}/{(2x3 + x2 - 5x + 7)}.
Solution:
We have,
Limx→∞{(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. Limx→∞{(5x3 - 6)}/{√(9 + 4x6)}.
Solution:
We have,
Limx→∞{(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. 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}
=
\lim_{x\to∞}\frac{1}{\sqrt{x}(1+\frac{1}{x}+1)} 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}
=
\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. 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)
=
\lim_{x\to∞}\frac{\sqrt{4+\frac{1}{x^2}}}{4} 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}
=
\lim_{n\to∞}\frac{n^2}{\frac{n(n+1)}{2}} = 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)
=
\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. 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
=
\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. 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
