Lim x -x2-1-x2-1-x4-1-x4-1 isequalto

The correct option is A 1 =x→∞lim√(x^2+1) √(x^2+1)/√(x^4+1) √(x^4 1) =x→∞limx√(1+1/x^2) x^2/3√(1+1/x^2)/x.√(1+1/x^4) x^4/5√(1 1/x^4) Dividing the numerator an ...

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Byju's Answer

Standard XI

Mathematics

Indeterminate Forms

lim x →∞√x2+1...

Question

$l i m x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{2} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} - 1}} i s e q u a l t o$

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Solution

The correct option is A
1

$= l i m x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{2} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} - 1}} = l i m x \to \infty \frac{x \sqrt{1 + \frac{1}{x^{2}}} - x^{2 / 3} \sqrt[3]{1 + \frac{1}{x^{2}}}}{x . \sqrt[4]{1 + \frac{1}{x^{4}}} - x^{4} / 5 \sqrt[5]{1 - \frac{1}{x^{4}}}} Dividing the numerator and denominator by x = l i m x \to \infty \frac{\sqrt{1 + \frac{1}{x^{2}}} - \frac{1}{x^{1} / 3} \sqrt[3]{1 + {\frac{1}{x}}^{2}}}{\sqrt[4]{1 + \frac{1}{x^{4}}} - \frac{1}{x^{1} / 5} \sqrt[5]{1 - \frac{1}{x^{4}}}} = \frac{1 - 0}{1 - 0} = 1$

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Similar questions

lim x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{2} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} - 1}}

is equal to

Q. Solve:

lim x \to \infty \frac{\sqrt{x^{2} + 1} - 3 \sqrt{x^{3} + 1}}{4 \sqrt{x^{4} + 1} + 5 \sqrt{x^{4} + 1}}

lim x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{3} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} + 1}}

is equals to

lim x \to 0 \frac{x (e^{(\sqrt{1 + x^{2} + x^{4}} - 1) / x} - 1)}{\sqrt{1 + x^{2} + x^{4}} - 1}

Solve

l i m x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{2} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} + 1}}

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limx→∞√x2+1−3√x2+14√x4+1−5√x4−1isequalto

The correct option is A 1 =limx→∞√x2+1−3√x2+14√x4+1−5√x4−1=limx→∞x√1+1x2−x2/33√1+1x2x.4√1+1x4−x4/55√1−1x4Dividing the numerator and denominator by x=limx→∞√1+1x2−1x1/33√1+1x24√1+1x4−1x1/55√1−1x4=1−01−0=1

$l i m x \to \infty \frac{\sqrt{x^{2} + 1} - \sqrt[3]{x^{2} + 1}}{\sqrt[4]{x^{4} + 1} - \sqrt[5]{x^{4} - 1}} i s e q u a l t o$