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Answer 1

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Answer 2

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Solution:-

Step-by-step explanation:

$Let \sqrt{arccos(x)} = t.$

Then we have.

$x = cos( {t}^{2} ).since \: x - > ( - 1) ^{ + }$

We have,

$= > {t}^{2} - > \pi ^{ - }$

Hence ,we have.

$</p><p>Lim_ > x( - 1) {}^{ + } \frac{ \sqrt{\pi - \ } \sqrt{ arccos(x)} }{ \sqrt{\pi - t} }$

$</p><p>Lim_ > t - > \sqrt{\pi {}^{ - } } \frac{ \sqrt{\pi - t} }{ \sqrt{1 + cos(( \sqrt{\pi - y)^{2}) } } }$

$1 + cos(( \sqrt{\pi} - {y})^{2} ) = 1 + cos$

$(\pi - 2 \sqrt{\pi \: y} + {y}^{2} )$

$= 1 - cos(2 \sqrt{\pi \: y} - {y}^{2} )$

$= 2sin {}^{2} ( \sqrt{\pi \: y} - \frac{y {}^{2} }{2} )$

Hence,

$lim > y - > 0 {}^{ + } \frac{y}{ \sqrt{1 + cos(( \sqrt{\pi \: - y) {}^{2}) } } }$

$= \frac{1}{ \sqrt{2} } lim > y - > 0 { }^{ + } \frac{y}{sin( \sqrt{\pi \: y - \frac{ {y}^{2} }{2} )} }$

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

$lim > 0 {}^{ + } \frac{( \sqrt{\pi \: y} - \frac{ {y}^{2} }{2} ) }{sin( \sqrt{\pi \: y - \frac{ {y}^{2} }{2} )( \sqrt{\pi \: y - \frac{ {y}^{2} }{2}) } } }$

$= \frac{1}{ \sqrt{2 \: \pi} }$

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