Question:

Step 1: Simplify $\sin(\cos^{-1} x)$ using a fundamental identity. Let $\theta = \cos^{-1} x$. Then $\cos \theta = x$, where $\theta \in [0, \pi]$. Using the Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$. Substituting $\cos \theta = x$, we get $\sin^2 \theta = 1 - x^2$. Since $\theta \in [0, \pi]$, $\sin \theta \ge 0$. Therefore, $\sin(\cos^{-1} x) = \sqrt{1 - x^2}$. Step 2: Rewrite the limit expression. The limit expression becomes: $$ \lim_{x \to -\frac{1}{\sqrt{2}}} \left( \sin(\cos^{-1} x) - x \right) = \lim_{x \to -\frac{1}{\sqrt{2}}} \left( \sqrt{1 - x^2} - x \right) $$ Step 3: Evaluate the limit by direct substitution. The function $f(x) = \sqrt{1 - x^2} - x$ is continuous on the interval $[-1, 1]$. Since $x = -\frac{1}{\sqrt{2}}$ is within this interval, the limit can be found by direct substitution. Step 4: Substitute $x = -\frac{1}{\sqrt{2}}$ into the expression. $$ f\left(-\frac{1}{\sqrt{2}}\right) = \sqrt{1 - \left(-\frac{1}{\sqrt{2}}\right)^2} - \left(-\frac{1}{\sqrt{2}}\right) $$ $$ = \sqrt{1 - \frac{1}{2}} + \frac{1}{\sqrt{2}} $$ $$ = \sqrt{\frac{1}{2}} + \frac{1}{\sqrt{2}} $$ $$ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} $$ $$ = \frac{2}{\sqrt{2}} $$ $$ = \frac{2\sqrt{2}}{2} $$ $$ = \sqrt{2} $$