(2) The range of the function $f(x)=\frac{\sqrt{x^{2}+1}}{x-1}$ is $\qquad$ .
(2) $\left(-\infty,-\frac{\sqrt{2}}{2}\right] \cup(1,+\infty)$ Hint: Let $x=\tan \theta, -\frac{\pi}{2} < \theta < \frac{\pi}{2}$, and $\theta \neq \frac{\pi}{4}$, then
$$
f(x)=\frac{\frac{1}{\cos \theta}}{\tan \theta-1}=\frac{1}{\sin \theta-\cos \theta}=\frac{1}{\sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)} .
$$

Let $u=\sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)$, then $-\sqrt{2} \leqslant u<1$, and $u \neq 0$, so $f(x)=$ $\frac{1}{u} \in\left(-\infty,-\frac{\sqrt{2}}{2}\right] \cup(1,+\infty)$