mathop {lim }limits_{x to 0} frac{{xsqrt {{y^ | vedclass

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(D) दिया गया सीमा: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$अंश क

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$\frac{1}{{128y}}$

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(D) दिया गया सीमा: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$अंश का सरलीकरण: $\sqrt {{y^2} - (y^2 - 2xy + x^2)} = \sqrt {2xy - x^2} = \sqrt x \sqrt {2y - x}$हर का सरलीकरण: $(\sqrt {4x(2y - x)} + \sqrt {8xy})^3 = (2\sqrt x \sqrt {2y - x} + 2\sqrt 2 \sqrt {xy})^3 = [2\sqrt x (\sqrt {2y - x} + \sqrt {2y})]^3 = 8x^{3/2}(\sqrt {2y - x} + \sqrt {2y})^3$सीमा में मान रखने पर:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{x \cdot \sqrt x \sqrt {2y - x} }}{{8x^{3/2}(\sqrt {2y - x} + \sqrt {2y})^3}}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{\sqrt {2y - x} }}{{8(\sqrt {2y - x} + \sqrt {2y})^3}}$जब $x 	o 0$:$L = \frac{{\sqrt {2y} }}{{8(\sqrt {2y} + \sqrt {2y})^3}} = \frac{{\sqrt {2y} }}{{8(2\sqrt {2y})^3}} = \frac{{\sqrt {2y} }}{{8 \cdot 8 \cdot 2\sqrt 2 \sqrt y \cdot y}} = \frac{{\sqrt 2 \sqrt y }}{{128 \sqrt 2 \sqrt y \cdot y}} = \frac{1}{{128y}}$

$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$ का मान ज्ञात कीजिए।

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