First you have to simplify the expression, start by multiplicating and dividing the whole expression by the conjugated of the denominator
$ \lim_ {x \to 1^{+}} \frac{x - 1}{\sqrt{2x - x^2} - 1} = \lim_ {x \to 1^{+}} \frac{x - 1}{\sqrt{2x - x^2} - 1} \times \frac{\sqrt{2x - x^2} + 1}{\sqrt{2x - x^2} + 1} $ $ = \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{(\sqrt{2x - x^2})^{2} - 1^{2}} = \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{-x^2 + 2x - 1} $
Notice that 1 is a root of the denominator, so we can factorate it using the Briot-Ruffini Method, and we get this
$ \lim_ {x \to 1^{+}} \frac{(x - 1) \times (\sqrt{2x - x^2} + 1)}{(x - 1) \times (-x + 1)} = $ $ \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{(-x + 1)} = $ $ \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{-(x -1)} = $ $ \lim_ {x \to 1^{+}} \frac{1}{-1} \times \frac{\sqrt{2x - x^2} + 1}{x -1} = $ $ -1 \times \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{x -1} $
Now let's verify for which values of x the expression x - 1 assumes positive values
$ x - 1 > 0 \leftrightarrow x > 1 $
As we are aproaching to x by values greater than 1, x - 1 aproachs to 0 by positive values, so
$ -1 \times \lim_ {x \to 1^{+}} \frac{\sqrt{2x - x^2} + 1}{x -1} = -1 \times \frac{2}{0^{+}} = -1 \times +\infty = -\infty $