1. Key Concepts and Formulas

• Limit of a function: Understanding how to evaluate limits as a variable approaches a certain value, especially when direct substitution leads to an indeterminate form.
• Algebraic Manipulation for Limits: Techniques like multiplying by the conjugate to simplify expressions and resolve indeterminate forms.
• Trigonometric Identities and Standard Limits: Knowledge of inverse trigonometric identities (e.g., $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$) and standard limits (e.g., $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$).
• Substitution in Limits: Replacing a variable with another to simplify the limit expression and change the limit point.

2. Step-by-Step Solution

The problem asks us to evaluate the limit: $L = \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }}$

• Step 1: Identify the Indeterminate Form. As $x \to 1^-$, $\sin^{-1}x \to \sin^{-1}(1) = \frac{\pi}{2}$. The numerator becomes $\sqrt{\pi} - \sqrt{2 \cdot \frac{\pi}{2}} = \sqrt{\pi} - \sqrt{\pi} = 0$. The denominator becomes $\sqrt{1-1} = \sqrt{0} = 0$. Thus, the limit is of the indeterminate form $\frac{0}{0}$.

• Step 2: Rationalize the Numerator. To handle the square roots in the numerator, we multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{\pi} + \sqrt{2\sin^{-1}x}$. $L = \mathop {\lim }\limits_{x \to {1^ - }} {{\sqrt \pi - \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt {1 - x} }} \times {{\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }}$ $L = \mathop {\lim }\limits_{x \to {1^ - }} {{(\sqrt \pi)^2 - (\sqrt {2{{\sin }^{ - 1}}x})^2 } \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$ $L = \mathop {\lim }\limits_{x \to {1^ - }} {{\pi - 2{{\sin }^{ - 1}}x} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

• Step 3: Utilize Inverse Trigonometric Identities. We can rewrite the term $\pi - 2\sin^{-1}x$ in the numerator. Recall the identity $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$. So, $\pi - 2\sin^{-1}x = 2\left(\frac{\pi}{2} - \sin^{-1}x\right) = 2\cos^{-1}x$. Substituting this into the limit expression: $L = \mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} \left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)}}$

• Step 4: Separate the Limit and Evaluate Parts. We can separate the limit into parts. The term $\left( {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} } \right)$ in the denominator approaches a finite, non-zero value as $x \to 1^-$. As $x \to 1^-$, $\sin^{-1}x \to \frac{\pi}{2}$. So, $\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} \to \sqrt \pi + \sqrt {2 \cdot \frac{\pi}{2}} = \sqrt \pi + \sqrt \pi = 2\sqrt \pi$. Therefore, we can write: $L = \left( \mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }} \right) \times \left( \mathop {\lim }\limits_{x \to {1^ - }} {1 \over {\sqrt \pi + \sqrt {2{{\sin }^{ - 1}}x} }} \right)$ $L = \left( \mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }} \right) \times \left( {1 \over {2\sqrt \pi }} \right)$

• Step 5: Evaluate the Remaining Limit using Substitution. Let's focus on the limit: $\mathop {\lim }\limits_{x \to {1^ - }} {{2{{\cos }^{ - 1}}x} \over {\sqrt {1 - x} }}$. We can use a substitution. Let $x = \cos \theta$. As $x \to 1^-$, $\cos \theta \to 1^-$. This implies $\theta \to 0^+$. Also, $\cos^{-1}x = \theta$. And $\sqrt{1-x} = \sqrt{1-\cos \theta}$. Using the half-angle identity $1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$, we get $\sqrt{1-\cos \theta} = \sqrt{2\sin^2(\frac{\theta}{2})} = \sqrt{2} |\sin(\frac{\theta}{2})|$. Since $\theta \to 0^+$, $\frac{\theta}{2} \to 0^+$, so $\sin(\frac{\theta}{2}) > 0$. Thus, $|\sin(\frac{\theta}{2})| = \sin(\frac{\theta}{2})$. So, $\sqrt{1-x} = \sqrt{2}\sin(\frac{\theta}{2})$. The limit becomes: $\mathop {\lim }\limits_{\theta \to {0^+}} {{2\theta } \over {\sqrt{2}\sin \left( {{\theta \over 2}} \right)}}$

• Step 6: Apply Standard Limit Formula. We can rewrite the expression to use the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$. $\mathop {\lim }\limits_{\theta \to {0^+}} {{2\theta } \over {\sqrt{2}\sin \left( {{\theta \over 2}} \right)}} = \mathop {\lim }\limits_{\theta \to {0^+}} {{2\theta } \over {\sqrt{2}}} \times {1 \over {\sin \left( {{\theta \over 2}} \right)}}$ $= \mathop {\lim }\limits_{\theta \to {0^+}} {{\sqrt{2}\theta } \over {\sin \left( {{\theta \over 2}} \right)}}$ To apply the standard limit, we need $\frac{\theta}{2}$ in the denominator. $= \mathop {\lim }\limits_{\theta \to {0^+}} \sqrt{2} \times {\theta \over {\sin \left( {{\theta \over 2}} \right)}} = \mathop {\lim }\limits_{\theta \to {0^+}} \sqrt{2} \times {2 \cdot \left( {\theta \over 2} \right) \over {\sin \left( {{\theta \over 2}} \right)}}$ $= \sqrt{2} \times 2 \times \mathop {\lim }\limits_{\theta \to {0^+}} {{\theta \over 2} \over {\sin \left( {{\theta \over 2}} \right)}}$ Let $y = \frac{\theta}{2}$. As $\theta \to 0^+$, $y \to 0^+$. $= 2\sqrt{2} \times \mathop {\lim }\limits_{y \to {0^+}} {y \over {\sin y}} = 2\sqrt{2} \times 1 = 2\sqrt{2}$

• Step 7: Combine the Results. Now, substitute the evaluated limit back into the expression for $L$: $L = (2\sqrt{2}) \times \left( {1 \over {2\sqrt \pi }} \right)$ $L = {{\sqrt{2}} \over {\sqrt \pi }}$ $L = \sqrt {{2 \over \pi }}$

3. Common Mistakes & Tips

• Incorrectly applying standard limits: Ensure the argument of the trigonometric function and the denominator match when using limits like $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$.
• Errors in trigonometric identities: Double-check identities like $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$ and $1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$.
• Substitution issues: Be careful with the direction of the limit ($x \to 1^-$ implies $\theta \to 0^+$) and the sign of the trigonometric function after substitution (e.g., $\sqrt{\sin^2(\frac{\theta}{2})}$).

4. Summary

The problem involves evaluating a limit that results in an indeterminate form $\frac{0}{0}$. We began by rationalizing the numerator to simplify the expression. Next, we used the inverse trigonometric identity $\frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$ to further simplify the numerator. By separating the limit and evaluating the part with the inverse trigonometric function using a substitution ($x = \cos \theta$) and a standard limit, we were able to find the value of the original limit.

The final answer is $\boxed{\sqrt {{2 \over \pi }}}$.