`lim_(xto 1^-) (sqrtpi-sqrt(2sin^-1x))/(sqrt(1-x))` is equal to

Options

A. `1/(sqrt(2pi))`
B. `sqrtpi/2`
C. `sqrt(2/pi)`
D. `sqrtpi`

Answer

To solve the limit \( \lim_{x \to 1^-} \frac{\sqrt{\pi} - \sqrt{2 \sin^{-1} x}}{\sqrt{1 - x}} \), we can apply L'Hôpital's Rule since we have an indeterminate form of type \( \frac{0}{0} \).

### Step 1: Identify the limit form
As \( x \to 1^- \):
- \( \sin^{-1}(1) = \frac{\pi}{2} \)
- Thus, \( \sqrt{2 \sin^{-1}(x)} \to \sqrt{2 \cdot \frac{\pi}{2}} = \sqrt{\pi} \)
- Therefore, the numerator \( \sqrt{\pi} - \sqrt{2 \sin^{-1} x} \to 0 \)
- The denominator \( \sqrt{1 - x} \to 0 \)

This confirms that we have the \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
We differentiate the numerator and denominator:

**Numerator:**
\[ \frac{d}{dx} (\sqrt{\pi} - \sqrt{2 \sin^{-1} x}) = 0 - \frac{1}{2\sqrt{2 \sin^{-1} x}} \cdot \frac{2}{\sqrt{1 - x^2}} = -\frac{1}{\sqrt{2 \sin^{-1} x} \cdot \sqrt{1 - x^2}} \]

**Denominator:**
\[ \frac{d}{dx} (\sqrt{1 - x}) = \frac{-1}{2\sqrt{1 - x}} \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 1^-} \frac{-\frac{1}{\sqrt{2 \sin^{-1} x} \cdot \sqrt{1 - x^2}}}{-\frac{1}{2\sqrt{1 - x}}} = \lim_{x \to 1^-} \frac{2\sqrt{1 - x}}{\sqrt{2 \sin^{-1} x} \cdot \sqrt{1 - x^2}} \]

### Step 4: Simplify the limit
As \( x \to 1^- \):
- \( \sin^{-1}(x) \to \frac{\pi}{2} \)
- \( \sqrt{2 \sin^{-1}(x)} \to \sqrt{\pi} \)
- \( \sqrt{1 - x} \to 0 \)
- \( \sqrt{1 - x^2} \to 0 \)

Thus, we can substitute these values into the limit:
\[ \lim_{x \to 1^-} \frac{2\sqrt{1 - x}}{\sqrt{\pi} \cdot 0} \to \text{undefined} \]

### Step 5: Re-evaluate using Taylor expansion
Instead of directly substituting, we can use the Taylor expansion around \( x = 1 \):
- \( \sin^{-1}(x) \approx \frac{\pi}{2} - \sqrt{2(1-x)} \)
- Thus, \( 2 \sin^{-1}(x) \approx \pi - 2\sqrt{2(1-x)} \)

Now substituting this back:
\[ \sqrt{2 \sin^{-1}(x)} \approx \sqrt{\pi - 2\sqrt{2(1-x)}} \]
As \( x \to 1^- \), we can find the limit:
\[ \sqrt{\pi} - \sqrt{2 \sin^{-1}(x)} \approx \sqrt{\pi} - \sqrt{\pi - 2\sqrt{2(1-x)}} \approx \frac{2\sqrt{2(1-x)}}{2\sqrt{\pi}} = \frac{\sqrt{2(1-x)}}{\sqrt{\pi}} \]

### Step 6: Substitute back into the limit
Now substituting back into our limit:
\[ \lim_{x \to 1^-} \frac{\frac{\sqrt{2(1-x)}}{\sqrt{\pi}}}{\sqrt{1-x}} = \lim_{x \to 1^-} \frac{\sqrt{2}}{\sqrt{\pi}} = \frac{\sqrt{2}}{\sqrt{\pi}} \]

### Final Answer
Thus, the limit evaluates to:
\[ \frac{\sqrt{2}}{\sqrt{\pi}} \]