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Finding $$lim_{xrightarrow 1^{-}}frac{sqrt{pi}-sqrt{2sin^{-1}(x)}}{sqrt{1-x}}$$

Try: put $h=1-x$

$$lim_{hrightarrow 0}frac{sqrt{pi}-sqrt{2sin^{-1}(1-h)}}{sqrt{h}}$$

D, L Hopital rule

$$lim_{hrightarrow 0}frac{sqrt{h}}{sqrt{2sin^{-1}(1-h)}}cdot frac{2}{sqrt{1-(1-h)^2}}=sqrt{frac{2}{pi}}$$

could some help me how to solve without D L Hopital Rule

$$lim_{xto1^-}dfrac{sqrtpi-sqrt{2sin^{-1}x}}{sqrt{1-x}}=lim_{xto1^-}dfrac1{sqrtpi+sqrt{2sin^{-1}x}}cdotlim_{xto1^-}dfrac{pi-2sin^{-1}x}{sqrt{1-x}}$$

Using Why it's true? $arcsin(x) +arccos(x) = frac{pi}{2}$

$$F=lim_{xto1^-}dfrac{pi-2sin^{-1}x}{sqrt{1-x}}=lim_{xto1^-}dfrac{2cos^{-1}x}{sqrt{1-x}}$$

Set $sqrt{1-x}=himplies x=1-h^2$

$$F=2lim_{hto0^+}dfrac{cos^{-1}(1-h^2)}h=2lim_{hto0^+}dfrac{sin^{-1}sqrt{1-(1-h^2)^2}}h$$

$$=2lim_{hto0^+}dfrac{sin^{-1}(hsqrt{2-h^2})}{hsqrt{2-h^2}}cdotlim_{hto0^+}sqrt{2-h^2}=?$$

If it is ok to use $lim_{uto 0}frac{sin u}{u} = 1$, then a possible way is setting $x=sin t$ and consider $tto frac{pi}{2}^-$:

begin{eqnarray*} frac{sqrt{pi}-sqrt{2sin^{-1}(x)}}{sqrt{1-x}}
& stackrel{x=sin t}{=} & frac{sqrt{pi}-sqrt{2t}}{sqrt{1-sin t}} \
& = & color{blue}{frac{sqrt{pi}-sqrt{2t}}{cos t}}cdot underbrace{sqrt{1+sin t}}_{stackrel{t to frac{pi}{2}^-}{longrightarrow}color{green}{sqrt{2}}} \
end{eqnarray*}

begin{eqnarray*} color{blue}{frac{sqrt{pi}-sqrt{2t}}{cos t}}
& = & color{orange}{frac{pi-2t}{cos t}} cdot underbrace{frac{1}{sqrt{pi} + sqrt{2t}}}_{stackrel{t to frac{pi}{2}^-}{longrightarrow}color{green}{frac{1}{2sqrt{pi}}}}\
end{eqnarray*}

begin{eqnarray*} color{orange}{frac{pi-2t}{cos t}}
& stackrel{t = u + frac{pi}{2}}{=} & underbrace{frac{-2u}{-sin u}}_{stackrel{u to 0^-}{longrightarrow}color{green}{2}}\
end{eqnarray*}

All together:
$$lim_{xrightarrow 1^{-}}frac{sqrt{pi}-sqrt{2sin^{-1}(x)}}{sqrt{1-x}} = color{green}{sqrt{2}cdot frac{1}{2sqrt{pi}} cdot 2} = boxed{frac{sqrt{2}}{sqrt{pi}}}$$

As you wrote $$y=frac{sqrt{pi}-sqrt{2sin^{-1}(1-h)}}{sqrt{h}}$$

Now, by Taylor series built at $h=0$
$$sin^{-1}(1-h)=frac{pi }{2}-sqrt{2} h^{1/2}-frac{h^{3/2}}{6 sqrt{2}}+Oleft(h^{5/2}right)$$
$$2sin^{-1}(1-h)=pi -2sqrt{2} h^{1/2}-frac{h^{3/2}}{3 sqrt{2}}+Oleft(h^{5/2}right)$$ Continuing with the binomial expansion
$$sqrt{2sin^{-1}(1-h) }=sqrt{pi }-sqrt{frac{2}{pi }} sqrt{h}-frac{h}{pi
^{3/2}}+Oleft(h^{3/2}right)$$ making
$$y=frac{sqrt{pi}-left(sqrt{pi }-sqrt{frac{2}{pi }} sqrt{h}-frac{h}{pi
^{3/2}}+Oleft(h^{3/2}right)right)}{sqrt{h}}=frac{sqrt{frac{2}{pi }} sqrt{h}+frac{h}{pi
^{3/2}}+Oleft(h^{3/2}right)}{sqrt{h}}$$ that is to say
$$y=sqrt{frac{2}{pi }} +frac{sqrt h}{pi
^{3/2}}+Oleft(hright)$$

If you substitute $y=sqrt{1-x}$, then $x=1-y^2$ and the limit becomes
$$
lim_{yto0^+}frac{sqrt{pi}-sqrt{2arcsinsqrt{1-y^2}}}{y}
$$
Not a simplification? Let's see. If $theta=arcsinsqrt{1-y^2}$, then $sqrt{1-y^2}=sintheta$ and so $y=costheta$ and $theta=arccos y$.

Thus your limit is
$$
lim_{yto0^+}frac{sqrt{pi}-sqrt{2arccos y}}{y}=
lim_{yto0^+}frac{2}{sqrt{pi}+sqrt{2arccos y}}frac{pi/2-arccos y}{y}=
lim_{yto0^+}frac{2}{sqrt{pi}+sqrt{2arccos y}}frac{arcsin y}{y}$$