$$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}$$ is equal to

We need to evaluate $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}$$.

Since $$\theta = \cos^{-1} x$$, as $$x \to \frac{1}{\sqrt{2}}$$, we have $$\theta \to \frac{\pi}{4}$$ and also $$\cos\theta = x$$ and $$\sin\theta = \sqrt{1 - x^2}$$. Now the given expression can be rewritten in terms of $$\theta$$ as $$\frac{\sin\theta - \cos\theta}{1 - \tan\theta}$$. Noting that $$1 - \tan\theta = 1 - \frac{\sin\theta}{\cos\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta}$$ gives $$\frac{\sin\theta - \cos\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} = (\sin\theta - \cos\theta)\,\frac{\cos\theta}{\cos\theta - \sin\theta}$$.

Substituting and simplifying yields $$\frac{(\sin\theta - \cos\theta)\cdot \cos\theta}{\cos\theta - \sin\theta} = \frac{-(\cos\theta - \sin\theta)\cdot \cos\theta}{\cos\theta - \sin\theta} = -\cos\theta$$. Next, taking the limit as $$\theta \to \frac{\pi}{4}$$ gives $$-\cos\frac{\pi}{4} = -\frac{1}{\sqrt{2}}$$.

The limit is $$-\frac{1}{\sqrt{2}}$$.

The correct answer is Option B.