mathop {lim }limits_{x to {1^ - }} frac{{sqrt | vedclass

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Question

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\sqrt \pi   - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$.અંશનું સંમેયીકરણ કરતા:$L

Vedclass · Accepted Answer

$\sqrt {\frac{2}{\pi }} $

Vedclass · Answer

(B) ધારો કે $L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\sqrt \pi   - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$.અંશનું સંમેયીકરણ કરતા:$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{\pi  - 2{{\sin }^{ - 1}}x}}{{\sqrt {1 - x} (\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}$$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{2(\frac{\pi }{2} - {{\sin }^{ - 1}}x)}}{{\sqrt {1 - x} (\sqrt \pi   + \sqrt {2{{\sin }^{ - 1}}x} )}}$$\frac{\pi }{2} - {{\sin }^{ - 1}}x = {{\cos }^{ - 1}}x$ નો ઉપયોગ કરતા:$L = \mathop {\lim }\limits_{x 	o {1^ - }} \frac{{2{{\cos }^{ - 1}}x}}{{\sqrt {1 - x} }} \cdot \frac{1}{{2\sqrt \pi  }}$$x = \cos 	heta$ લેતા,જ્યારે $x 	o 1^-$,ત્યારે $	heta 	o 0^+$. $\sqrt{1-x} = \sqrt{2}\sin(	heta/2)$.$L = \mathop {\lim }\limits_{	heta 	o 0^+} \frac{{2	heta }}{{\sqrt{2}\sin(	heta/2)}} \cdot \frac{1}{{2\sqrt \pi  }} = \sqrt{\frac{2}{\pi}}$.

$\mathop {\lim }\limits_{x \to {1^ - }} \frac{{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} }}{{\sqrt {1 - x} }}$ ની કિંમત શોધો.

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