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
$$\lim_{x \rightarrow (-1)^{+}}\left(\frac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}} \right)$$

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}
