5. The range of real numbers $x$ that satisfy $\sqrt{1-x^{2}} \geqslant x$ is $\qquad$
5. $\left[-1, \frac{\sqrt{2}}{2}\right]$.

Let $y=\sqrt{1-x^{2}}$, this is the upper half of the unit circle, which intersects the line $y=x$ at the point $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. The portion of the semicircle to the left of the intersection point lies above the line $y=x$.
Therefore, $x \in\left[-1, \frac{\sqrt{2}}{2}\right]$.