1. We are given that $\tan x = \frac{1}{\sqrt{3}}$ and $0^\circ \leq x \leq 90^\circ$. 2. Recall that $\tan x = \frac{\sin x}{\cos x}$, so we have $$\frac{\sin x}{\cos x} = \frac{1}{\sqrt{3}}.$$ 3. This implies $$\sin x = \frac{\cos x}{\sqrt{3}}.$$ 4. We need to find $\cos x - \sin x$. Substitute $\sin x$: $$\cos x - \sin x = \cos x - \frac{\cos x}{\sqrt{3}} = \cos x \left(1 - \frac{1}{\sqrt{3}}\right).$$ 5. Simplify the factor: $$1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}.$$ 6. To find $\cos x$, use $\sin^2 x + \cos^2 x = 1$: $$\left(\frac{\cos x}{\sqrt{3}}\right)^2 + \cos^2 x = 1,$$ $$\frac{\cos^2 x}{3} + \cos^2 x = 1,$$ $$\cos^2 x \left(\frac{1}{3} + 1\right) = 1,$$ $$\cos^2 x \cdot \frac{4}{3} = 1,$$ $$\cos^2 x = \frac{3}{4},$$ so $$\cos x = \frac{\sqrt{3}}{2}$$ since $x$ is in the first quadrant. 7. Substitute back: $$\cos x - \sin x = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3} - 1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{2}.$$ **Final answer:** $$\cos x - \sin x = \frac{\sqrt{3} - 1}{2}.$$