$$\lim_{x \to 1^-} \frac{\sqrt{\pi} - \sqrt{2\sin^{-1}x}}{\sqrt{1-x}}$$ is equal to

We have to evaluate the limit

$$$\lim_{x \to 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}}.$$$

First observe what happens to the numerator and the denominator separately as $$x$$ approaches $$1$$ from the left.

When $$x \to 1^{-}$$, we know that $$\sin^{-1}x$$ (that is, $$\arcsin x$$) approaches $$\dfrac{\pi}{2}$$. Therefore

$$$\sqrt{\pi}-\sqrt{2\sin^{-1}x}\; \longrightarrow\; \sqrt{\pi}-\sqrt{2\cdot\frac{\pi}{2}} = \sqrt{\pi}-\sqrt{\pi}=0.$$$

Simultaneously, the denominator also approaches zero, because

$$\sqrt{1-x}\; \longrightarrow\; \sqrt{1-1}=0.$$

So the limit has the indeterminate form $$\dfrac{0}{0}$$. Whenever we face a $$\dfrac{0}{0}$$ form, L’Hospital’s Rule is applicable. The rule states:

If $$\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$$ and the derivatives exist near $$a$$, then $$$\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$$.

So we differentiate the numerator and the denominator with respect to $$x$$.

Derivative of the numerator

The numerator is $$f(x)=\sqrt{\pi}-\sqrt{2\sin^{-1}x}$$. The constant $$\sqrt{\pi}$$ differentiates to $$0$$, so we focus on the second term.

First recall that for any differentiable function $$u(x)$$,

$$$\dfrac{d}{dx}\sqrt{u(x)}=\dfrac{1}{2\sqrt{u(x)}}\,u'(x).$$$

Here $$u(x)=2\sin^{-1}x$$. Thus,

$$$f'(x)=-\dfrac{1}{2\sqrt{2\sin^{-1}x}}\;\dfrac{d}{dx}\bigl(2\sin^{-1}x\bigr).$$$

Now $$$\dfrac{d}{dx}\bigl(2\sin^{-1}x\bigr)=2\cdot\dfrac{1}{\sqrt{1-x^{2}}}$$$ because the derivative of $$\sin^{-1}x$$ is $$\dfrac{1}{\sqrt{1-x^{2}}}$$.

Substituting this derivative back, we obtain

$$$f'(x) =-\dfrac{1}{2\sqrt{2\sin^{-1}x}}\;\Bigl(2\cdot\dfrac{1}{\sqrt{1-x^{2}}}\Bigr) =-\dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}.$$$

Derivative of the denominator

The denominator is $$g(x)=\sqrt{1-x}$$. Using the same square-root derivative formula with $$v(x)=1-x$$, we get

$$$g'(x)=\dfrac{1}{2\sqrt{1-x}}\;\dfrac{d}{dx}(1-x) =\dfrac{1}{2\sqrt{1-x}}\;(-1) =-\dfrac{1}{2\sqrt{1-x}}.$$$

Applying L’Hospital’s Rule

Therefore,

$$$\lim_{x \to 1^-}\dfrac{\sqrt{\pi}-\sqrt{2\sin^{-1}x}}{\sqrt{1-x}} =\lim_{x \to 1^-}\dfrac{f'(x)}{g'(x)} =\lim_{x \to 1^-} \dfrac{-\dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}} {-\dfrac{1}{2\sqrt{1-x}}}. $$$

The negatives cancel, giving

$$$\lim_{x \to 1^-} \dfrac{1}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}} \;\bigg/\; \dfrac{1}{2\sqrt{1-x}} =\lim_{x \to 1^-} \dfrac{2\sqrt{1-x}}{\sqrt{2\sin^{-1}x}\,\sqrt{1-x^{2}}}. $$$

We now simplify the square roots in the denominator. Notice that

$$1-x^{2}=(1-x)(1+x),$$

so

$$$\sqrt{1-x^{2}}=\sqrt{(1-x)(1+x)}=\sqrt{1-x}\,\sqrt{1+x}.$$$

Substituting this factorisation, we obtain

$$$\dfrac{2\sqrt{1-x}} {\sqrt{2\sin^{-1}x}\,\bigl(\sqrt{1-x}\,\sqrt{1+x}\bigr)} =\dfrac{2}{\sqrt{2\sin^{-1}x}\,\sqrt{1+x}}.$$$

At this stage the factor $$\sqrt{1-x}$$ cancels completely, so the limit is no longer indeterminate. We can now let $$x$$ approach $$1$$:

When $$x \to 1^{-}$$,

$$\sin^{-1}x \to \dfrac{\pi}{2}$$  and  $$1+x \to 2.$$

Hence

$$$\sqrt{2\sin^{-1}x}\; \longrightarrow\; \sqrt{2\cdot\dfrac{\pi}{2}}=\sqrt{\pi},$$$

and

$$\sqrt{1+x}\;\longrightarrow\;\sqrt{2}.$$

Therefore the limit equals

$$$\dfrac{2}{\sqrt{\pi}\,\sqrt{2}} =\dfrac{2}{\sqrt{2}\,\sqrt{\pi}} =\left(\dfrac{2}{\sqrt{2}}\right)\,\dfrac{1}{\sqrt{\pi}} =\sqrt{2}\,\dfrac{1}{\sqrt{\pi}} =\sqrt{\dfrac{2}{\pi}}.$$$

This value corresponds to Option B.

Hence, the correct answer is Option B.