lim _{x rightarrow infty} x^3 left[ sqrt{x^2 | vedclass

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Question

(C) ધારો કે $L = \lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right]$.પદનું સંમેયીકરણ કરતા:$L = \lim _{x \rightar

Vedclass · Accepted Answer

$\frac{1}{4 \sqrt{2}}$

Vedclass · Answer

(C) ધારો કે $L = \lim _{x ightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x ight]$.પદનું સંમેયીકરણ કરતા:$L = \lim _{x ightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x ight] 	imes \frac{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$$= \lim _{x ightarrow \infty} x^3 \frac{x^2 + \sqrt{x^4 + 1} - 2x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x} = \lim _{x ightarrow \infty} x^3 \frac{\sqrt{x^4 + 1} - x^2}{\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x}$.ફરીથી સંમેયીકરણ કરતા:$= \lim _{x ightarrow \infty} x^3 \frac{(\sqrt{x^4 + 1} - x^2)(\sqrt{x^4 + 1} + x^2)}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)} = \lim _{x ightarrow \infty} x^3 \frac{x^4 + 1 - x^4}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$$= \lim _{x ightarrow \infty} \frac{x^3}{(\sqrt{x^2 + \sqrt{x^4 + 1}} + \sqrt{2} x)(\sqrt{x^4 + 1} + x^2)}$.અંશ અને છેદને $x^3$ વડે ભાગતા:$= \lim _{x ightarrow \infty} \frac{1}{(\sqrt{1 + \sqrt{1 + \frac{1}{x^4}}} + \sqrt{2})(\sqrt{1 + \frac{1}{x^4}} + 1)} = \frac{1}{(\sqrt{1 + 1} + \sqrt{2})(1 + 1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.

$\lim _{x \rightarrow \infty} x^3 \left[ \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right] = $

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