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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) આપેલ લક્ષ: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$અંશનું સા

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) આપેલ લક્ષ: $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$અંશનું સાદું રૂપ: $\sqrt {{y^2} - (y^2 - 2xy + x^2)} = \sqrt {2xy - x^2} = \sqrt x \sqrt {2y - x}$છેદનું સાદું રૂપ: $(\sqrt {4x(2y - x)} + \sqrt {8xy})^3 = (2\sqrt x \sqrt {2y - x} + 2\sqrt 2 \sqrt {xy})^3 = [2\sqrt x (\sqrt {2y - x} + \sqrt {2y})]^3 = 8x^{3/2}(\sqrt {2y - x} + \sqrt {2y})^3$લક્ષમાં કિંમત મૂકતા:$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}}$જ્યારે $x 	o 0$:$L = \frac{{\sqrt {2y} }}{{8(\sqrt {2y} + \sqrt {2y})^3}} = \frac{{\sqrt {2y} }}{{8(2\sqrt {2y})^3}} = \frac{{\sqrt {2y} }}{{8 \cdot 8 \cdot 2\sqrt 2 \sqrt y \cdot y}} = \frac{{\sqrt 2 \sqrt y }}{{128 \sqrt 2 \sqrt y \cdot y}} = \frac{1}{{128y}}$

$\mathop {\lim }\limits_{x \to 0} \frac{{x\sqrt {{y^2} - {{(y - x)}^2}} }}{{{{(\sqrt {8xy - 4{x^2}} + \sqrt {8xy} )}^3}}}$ ની કિંમત શોધો.

Similar Questions

લક્ષની કિંમત શોધો: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$

$\mathop {\lim }\limits_{x \to 2} \frac{|x - 2|}{x - 2} = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $

$\lim\limits_{x \rightarrow 2} \frac{3^{x}+3^{3-x}-12}{3^{-x / 2}-3^{1-x}}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module