13. Let the function $f(x)=x-k \sqrt{x^{2}-1}(x \geqslant 1$, $k$ is a given real number, $0<k<1)$. Try to find the range of $f(x)$.
Three, 13. When $x>1$, the derivative of $f(x)$ is
$f^{\prime}(x)=1-\frac{k x}{\sqrt{x^{2}-1}}$.
Let $f^{\prime}(t)=0$.
When $t>1$, the solution is $t=\frac{1}{\sqrt{1-k^{2}}}$.
Thus, $f(t)=f\left(\frac{1}{\sqrt{1-k^{2}}}\right)=\sqrt{1-k^{2}}$, $f(1)=1$.
Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline$x$ & 1 & $(1, t)$ & $t$ & $(t,+\infty)$ \\
\hline$f^{\prime}(x)$ & & - & 0 & + \\
\hline$f(x)$ & 1 & decreasing & minimum & increasing \\
\hline
\end{tabular}

When $x \rightarrow+\infty$,
$$
f(x)=\frac{\left(1-k^{2}\right) x^{2}+k^{2}}{x+k \sqrt{x^{2}-1}} \rightarrow+\infty \text {. }
$$

Therefore, the range of $f(x)$ is $\left[\sqrt{1-k^{2}},+\infty\right)$.