Calculate:$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$ How to calculate following with out using L'Hospital rule
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$
 A: Let $\sqrt{\arccos(x)} = t$. We then have $x = \cos(t^2)$. Since $x \to (-1)^+$, we have $t^2 \to \pi^-$. Hence, we have
$$\lim_{x \to (-1)^+} \dfrac{\sqrt{\pi} - \sqrt{\arccos(x)}}{\sqrt{1+x}} = \overbrace{\lim_{t \to \sqrt{\pi}^-} \dfrac{\sqrt{\pi} - t}{\sqrt{1+\cos(t^2)}}}^{t = \sqrt{\arccos(x)}} = \underbrace{\lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}}}_{y = \sqrt{\pi}-t}$$
$$1+\cos((\sqrt{\pi}-y)^2) = 1+\cos(\pi -2 \sqrt{\pi}y + y^2) = 1-\cos(2 \sqrt{\pi}y - y^2) = 2 \sin^2 \left(\sqrt{\pi} y - \dfrac{y^2}2\right)$$
Hence,
\begin{align}
\lim_{y \to 0^+} \dfrac{y}{\sqrt{1+\cos((\sqrt{\pi}-y)^2)}} & = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{y}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)}\\
& = \dfrac1{\sqrt2} \lim_{y \to 0^+} \dfrac{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)}{\sin \left(\sqrt{\pi}y - \dfrac{y^2}2\right)} \dfrac{y}{\left(\sqrt{\pi}y - \dfrac{y^2}2\right)} = \dfrac1{\sqrt{2 \pi}}
\end{align}
A: \begin{eqnarray*}
\lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{\pi}-\sqrt{\arccos x}}{\sqrt{x+1}} & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\pi-\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\
 & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\left(\pi-\arccos x\right)}{2\sqrt{\pi}\sqrt{x+1}}\\
 & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sin\arccos x}{2\sqrt{\pi}\sqrt{x+1}}\\
 & = & \lim_{x\rightarrow\left(-1\right)^{+}}\frac{\sqrt{1-x^{2}}}{2\sqrt{\pi}\sqrt{x+1}}\\
 & = & \frac{1}{\sqrt{2\pi}}
\end{eqnarray*}
A: Provided the following limits exist,
\begin{align*}
A &= \lim_{x \to -1^{+}}\left(\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{\sqrt{x+1}} \right)\\
&= \lim_{x\to -1^+}\left(\frac{\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\frac{\sqrt{x+1} - 0}{x + 1}} \right)\\
&= \frac{\lim_{x\to -1^+}\frac{\sqrt{\pi}-\sqrt{cos^{-1}x}}{x + 1}}{\lim_{x\to -1^+}\frac{\sqrt{x+1} - 0}{x + 1}},
\end{align*}
which we recognize as the quotient of the "derivative from the right" of $\sqrt{\cos^{-1} x}$ and $\sqrt{ x + 1}$ at $x = -1$. So,
\begin{align*}
A &= \lim_{x\to -1^+}\frac{\frac{1}{2\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}}{\frac{1}{2\sqrt{x + 1}}}\\
&= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x^2}\sqrt{\cos^{-1} x}}\\
&= \lim_{x\to -1^+}\frac{\sqrt{x + 1}}{\sqrt{1 - x}\sqrt{1 + x}\sqrt{\cos^{-1} x}}\\
&= \lim_{x\to -1^+}\frac{1}{\sqrt{1 - x}\sqrt{\cos^{-1} x}}\\
&= \frac{1}{\sqrt{2\pi}}.\\
\end{align*}
(in the first equation we flip the sign of the derivative because they're being taken in different orders: the top has the value first, limit second, bottom has limit first, value second.) 
Admittedly, I'm playing pretty fast and loose here, but if you're ambitious and have a bit of tenacity (and free time), you can probably fill in the missing details.
A: Hint
$$ \lim_{x \to (-1)^+} \cos^{-1} x = \lim_{ h \to 0 } {\cos^{-1} (-1 + h) }$$
From here, you can follow this previous question.
A: \begin{align}
L &=\lim_{x \rightarrow (-1)^{+}}\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}\\
&=\lim_{t \rightarrow 0^{+}}\frac{\sqrt{\pi}-\sqrt{\cos^{-1}(t-1)}}{\sqrt{t}}\\
&= \lim_{t \rightarrow 0^{+}}\frac{\sqrt{\pi}-\sqrt{\pi}+\frac{\sqrt{t}}{\sqrt{2\pi}}+o(t)}{\sqrt{t}}\\
&= \frac{1}{\sqrt{2\pi}}
\end{align}
