13. Given the function $f(x)=x-k \sqrt{x^{2}-1}(x \geqslant 1)$, where $k$ is a given real number, $0<k<1$. Try to find the range of $f(x)$.
(13) Method 1: When $x>1$, the derivative of $f(x)$ is
$$
f^{\prime}(x)=1-\frac{k x}{\sqrt{x^{2}-1}} \text {. }
$$

Let $f^{\prime}(t)=0$, since $t>1$, solving gives $t=\frac{1}{\sqrt{1-k^{2}}}$. Therefore,
$$
\begin{aligned}
f(t) & =f\left(\frac{1}{\sqrt{1-k^{2}}}\right) \\
& =\frac{1}{\sqrt{1-k^{2}}}-\frac{k^{2}}{\sqrt{1-k^{2}}} \\
& =\sqrt{1-k^{2}},
\end{aligned}
$$
$f(1)=1$. The table is as follows:
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & 1 & $(1, t)$ & $t$ & $(t,+\infty)$ \\
\hline$f^{\prime}(x)$ & & - & 0 & + \\
\hline$f(x)$ & 1 & $\downarrow$ & minimum & $\nearrow$ \\
\hline
\end{tabular}

Method 2: Let $x=\sec \theta, \theta \in\left[0, \frac{\pi}{2}\right)$, then $\sqrt{x^{2}-1}=\tan \theta$. Let
$$
f(x)=u=\sec \theta-k \tan \theta=\frac{1-k \sin \theta}{\cos \theta},
$$

we know
$$
u \cos \theta+k \sin \theta=1,
$$

then
$$
\sin (\theta+\varphi)=\frac{1}{\sqrt{u^{2}+k^{2}}}
$$

where
$$
\sin \varphi=\frac{u}{\sqrt{u^{2}+k^{2}}}, \cos \varphi=\frac{k}{\sqrt{u^{2}+k^{2}}} .
$$

Since $u>0$, by $|\sin \theta| \leqslant 1$, we get $u^{2} \geqslant 1-k^{2}$, i.e., $u \geqslant \sqrt{1-k^{2}}$. For all $u$ values not less than $\sqrt{1-k^{2}}$, we have $\frac{1}{\sqrt{u^{2}+k^{2}}} \leqslant 1$, thus there exist $\varphi$ and $\theta$ such that
$$
\begin{array}{c}
\sin \varphi=\frac{u}{\sqrt{u^{2}+k^{2}}}, \\
\cos \varphi=\frac{k}{\sqrt{u^{2}+k^{2}}}, \\
\sin (\theta+\varphi)=\frac{1}{\sqrt{u^{2}+k^{2}}}
\end{array}
$$

hold. Thus $u=\sec \theta-k \tan \theta$, i.e., there exists $x=\sec \theta$ such that $x-k \sqrt{x^{2}-1}=u$ holds. Therefore, the range sought is $\left[\sqrt{1-k^{2}},+\infty\right)$.