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

Try: put $h=1-x$

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

D, L Hopital rule

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

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

DXT
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    It's very tempting to try to make this into the definition of the derivative of $2\sin^{-1}(x)$ (either by multiplying with $\frac{\sqrt{1-x}}{\sqrt{1-x}}$ or by squaring the fraction), but personally, I can't make that happen without doing questionable limit manipulations. – Arthur Jan 15 '19 at 07:42

4 Answers4

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

\begin{eqnarray*} \frac{\sqrt{\pi}-\sqrt{2\sin^{-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}{2\sqrt{\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: $$\lim_{x\rightarrow 1^{-}}\frac{\sqrt{\pi}-\sqrt{2\sin^{-1}(x)}}{\sqrt{1-x}} = \color{green}{\sqrt{2}\cdot \frac{1}{2\sqrt{\pi}} \cdot 2} = \boxed{\frac{\sqrt{2}}{\sqrt{\pi}}}$$

trancelocation
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$$\lim_{x\to1^-}\dfrac{\sqrt\pi-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}=\lim_{x\to1^-}\dfrac1{\sqrt\pi+\sqrt{2\sin^{-1}x}}\cdot\lim_{x\to1^-}\dfrac{\pi-2\sin^{-1}x}{\sqrt{1-x}}$$

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

$$F=\lim_{x\to1^-}\dfrac{\pi-2\sin^{-1}x}{\sqrt{1-x}}=\lim_{x\to1^-}\dfrac{2\cos^{-1}x}{\sqrt{1-x}}$$

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

$$F=2\lim_{h\to0^+}\dfrac{\cos^{-1}(1-h^2)}h=2\lim_{h\to0^+}\dfrac{\sin^{-1}\sqrt{1-(1-h^2)^2}}h$$

$$=2\lim_{h\to0^+}\dfrac{\sin^{-1}(h\sqrt{2-h^2})}{h\sqrt{2-h^2}}\cdot\lim_{h\to0^+}\sqrt{2-h^2}=?$$

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

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

Claude Leibovici
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If you substitute $y=\sqrt{1-x}$, then $x=1-y^2$ and the limit becomes $$ \lim_{y\to0^+}\frac{\sqrt{\pi}-\sqrt{2\arcsin\sqrt{1-y^2}}}{y} $$ Not a simplification? Let's see. If $\theta=\arcsin\sqrt{1-y^2}$, then $\sqrt{1-y^2}=\sin\theta$ and so $y=\cos\theta$ and $\theta=\arccos y$.

Thus your limit is $$ \lim_{y\to0^+}\frac{\sqrt{\pi}-\sqrt{2\arccos y}}{y}= \lim_{y\to0^+}\frac{2}{\sqrt{\pi}+\sqrt{2\arccos y}}\frac{\pi/2-\arccos y}{y}= \lim_{y\to0^+}\frac{2}{\sqrt{\pi}+\sqrt{2\arccos y}}\frac{\arcsin y}{y}$$

egreg
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