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

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.
