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Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\lim_{x\to1^{-}}\,\dfrac{\sqrt{\pi}-\sqrt{2\,sin^{-1}(x)}}{\sqrt{1-x}}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{\sqrt{\pi}-\sqrt{2\,sin^{-1}(1-h)}}{\sqrt{1-(1-h)}}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{\sqrt{\pi}-\sqrt{2\,sin^{-1}(1-h)}}{\sqrt{h}}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{\left(\sqrt{\pi}-\sqrt{2\,sin^{-1}(1-h)}\right)\left(\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}\right)}{\sqrt{h}\cdot\,\left(\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}\right)}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{\pi-2\,sin^{-1}(1-h)}{\sqrt{h}\cdot\,\left(\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}\right)}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{2\left\{\dfrac{\pi}{2}-\,sin^{-1}(1-h)\right\}}{\sqrt{h}\cdot\,\left(\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}\right)}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{2\,cos^{-1}(1-h)}{\sqrt{h}\cdot\,\left(\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}\right)}}$

$\tt{=\displaystyle\lim_{h\to0}\,\dfrac{2\,cos^{-1}(1-h)}{\sqrt{h}}\cdot\,\lim_{h\to0}\dfrac{1}{\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-h)}}}$

$\tt{=2\,\displaystyle\lim_{h\to0}\,\dfrac{cos^{-1}(1-h)}{\sqrt{h}}\cdot\,\dfrac{1}{\sqrt{\pi}+\sqrt{2\,sin^{-1}(1-0)}}}$

$\tt{=2\,\displaystyle\lim_{h\to0}\,\dfrac{cos^{-1}(1-h)}{\sqrt{h}}\cdot\,\dfrac{1}{\sqrt{\pi}+\sqrt{2\,sin^{-1}(1)}}}$

$\tt{=2\,\displaystyle\lim_{h\to0}\,\dfrac{cos^{-1}(1-h)}{\sqrt{h}}\cdot\,\dfrac{1}{\sqrt{\pi}+\sqrt{\pi}}}$

$\tt{=2\,\displaystyle\lim_{h\to0}\,\dfrac{cos^{-1}(1-h)}{\sqrt{h}}\cdot\,\dfrac{1}{2\sqrt{\pi}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{cos^{-1}(1-h)}{\sqrt{h}}}$

Using l'hospital's rule,

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{\dfrac{-1}{\sqrt{1-(1-h)^2}}}{\dfrac{1}{2\sqrt{h}}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2\sqrt{h}}{\sqrt{1-(1-h)^2}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2\sqrt{h}}{\sqrt{1-(1-2h+h^2)}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2\sqrt{h}}{\sqrt{1-1+2h-h^2}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2\sqrt{h}}{\sqrt{2h-h^2}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2\sqrt{h}}{\sqrt{h}\,\sqrt{2-h}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\,\displaystyle\lim_{h\to0}\,\dfrac{-2}{\sqrt{2-h}}}$

$\tt{=\dfrac{1}{\sqrt{\pi}}\cdot\dfrac{-2}{\sqrt{2-0}}}$

$\tt{=-\dfrac{1}{\sqrt{\pi}}\cdot\dfrac{2}{\sqrt{2}}}$

$\tt{=-\dfrac{1}{\sqrt{\pi}}\cdot\sqrt{2}}$

$\tt{=-\sqrt{\dfrac{2}{\pi}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1^-} \frac{ \sqrt{\pi} - \sqrt{ {2sin}^{ - 1}x } }{ \sqrt{1 - x} }$

On substituting,

$\boxed{ \tt{ \: x = 1 - h, \: \: as \: x \to \: 1 \: \: so \: h \: \to \: 0 \: }}$

So, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{ \sqrt{\pi} - \sqrt{ {2sin}^{ - 1}(1 - h)} }{ \sqrt{1 - (1 - h)} }$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{ \sqrt{\pi} - \sqrt{ {2sin}^{ - 1}(1 - h)} }{ \sqrt{1 - 1 + h} }$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{ \sqrt{\pi} - \sqrt{ {2sin}^{ - 1}(1 - h)} }{ \sqrt{h} }$

Now, Using L - Hospital Rule, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{\dfrac{d}{dh} \sqrt{\pi} - \dfrac{d}{dh} \sqrt{ {2sin}^{ - 1}(1 - h)} }{ \dfrac{d}{dh}\sqrt{h} }$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{0 - \dfrac{1}{2 \sqrt{2 {sin}^{ - 1} (1 - h)}}\dfrac{d}{dh}2 {sin}^{ - 1}(1 - h) }{\dfrac{1}{2 \sqrt{h} } }$

$\rm \:  = - 2\displaystyle\lim_{h \to 0} \frac{\dfrac{1}{\sqrt{2 {sin}^{ - 1} (1 - h)}}\dfrac{1}{ \sqrt{1 - {(1 - h)}^{2}} }( - 1) }{\dfrac{1}{\sqrt{h} } }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ \sqrt{h} }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{1 - (1 + {h}^{2} - 2h) } }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ \sqrt{h} }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{1 - 1 - {h}^{2} + 2h } }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ \sqrt{h} }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{ - {h}^{2} + 2h } }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ \sqrt{h} }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{h(2 - h)} }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ \sqrt{h} }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{h} \sqrt{(2 - h)} }$

$\rm \:  =  \: 2\displaystyle\lim_{h \to 0} \frac{ 1 }{ \sqrt{2 {sin}^{ - 1}(1 - h) } \sqrt{(2 - h)} }$

$\rm \:  =  \: \dfrac{2}{ \sqrt{2} \times \sqrt{2} \times \sqrt{ {sin}^{ - 1} 1} }$

$\rm \:  =  \: \dfrac{1}{\sqrt{\dfrac{\pi}{2} }}$

$\rm \:  =  \: \sqrt{\dfrac{2}{\pi} }$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 1^-} \frac{ \sqrt{\pi} - \sqrt{ {2sin}^{ - 1}x } }{ \sqrt{1 - x} } = \sqrt{\dfrac{2}{\pi} } \: }}}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$