The point $x\=1$ is the left endpoint of the domain of $f$, and at this point, $f$ is discontinuous; the discontinuity is not removable. Here's how these conclusions are reached.

The graph of $f$ in Figure 1.6.4(a) suggests its domain is the open interval $\left(1\,\infty \right)$.

Since $f\left(1\right)$ is undefined (division by zero), $x\&equals;1$ is not in the domain of $f$, and likewise any $x<1$, but these because the square root is real only where $x-1\&gt;0$.

Because $f\left(1\right)$ is undefined, $f$ cannot be continuous at $x\&equals;1$.

Since $x\&equals;1$ is the equation of a vertical asymptote for $f$  (because $\underset{x\to 1\&plus;}{lim}1\&sol;\sqrt{x-1}$ = $\mathrm{\&infin;}$), $f\left(1\right)$ cannot be defined to make its extension to $\left[1\&comma;\infty \right]$ continuous.

A function cannot be continuous at a point where there is a vertical asymptote.

Such a discontinuity is not removable.