यदि a = lim_{x rightarrow 0} frac{sqrt{1+sqrt | vedclass

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(B) के लिए: $a = \lim_{x ightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} 	imes \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) के लिए: $a = \lim_{x ightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4} 	imes \frac{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+x^4}}+\sqrt{2}} = \lim_{x ightarrow 0} \frac{\sqrt{1+x^4}-1}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})} = \lim_{x ightarrow 0} \frac{x^4}{x^4(\sqrt{1+\sqrt{1+x^4}}+\sqrt{2})(\sqrt{1+x^4}+1)} = \frac{1}{(2\sqrt{2})(2)} = \frac{1}{4\sqrt{2}}$.$b$ के लिए: $b = \lim_{x ightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}} = \lim_{x ightarrow 0} \frac{(1-\cos^2 x)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} = \lim_{x ightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x} = \lim_{x ightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x}) = (1+1)(\sqrt{2}+\sqrt{2}) = 2 	imes 2\sqrt{2} = 4\sqrt{2}$.$ab^3$ की गणना: $ab^3 = \frac{1}{4\sqrt{2}} 	imes (4\sqrt{2})^3 = \frac{1}{4\sqrt{2}} 	imes 64 	imes 2\sqrt{2} = \frac{128\sqrt{2}}{4\sqrt{2}} = 32$.

यदि $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ और $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ है,तो $ab^3$ का मान ज्ञात कीजिए।

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