To evaluate the limit

$\lim_{{x \to 1}} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2 + 2x}}$

we’ll start by directly substituting $x = 1$ into the expression to see if it results in an indeterminate form:

1. Substitute $x = 1$:

   $\frac{1 - \sqrt{2 - 1^2}}{2 \cdot 1 - \sqrt{2 + 2 \cdot 1}} = \frac{1 - \sqrt{2 - 1}}{2 - \sqrt{2 + 2}}$

   Simplifying inside the square roots:

   $= \frac{1 - \sqrt{1}}{2 - \sqrt{4}} = \frac{1 - 1}{2 - 2} = \frac{0}{0}$

   This results in the indeterminate form $\frac{0}{0}$, so we need to manipulate the expression further to evaluate the limit.

2. Rationalize the numerator and denominator:

   To resolve the indeterminate form, we can multiply both the numerator and the denominator by the conjugate expressions of each part.

   Rationalize the numerator: Multiply the numerator and the denominator by the conjugate of $x - \sqrt{2 - x^2}$, which is $x + \sqrt{2 - x^2}$:

   $\frac{\left( x - \sqrt{2 - x^2} \right)\left( x + \sqrt{2 - x^2} \right)}{\left( 2x - \sqrt{2 + 2x} \right) \left( x + \sqrt{2 - x^2} \right)}$

   This simplifies the numerator to:

   $(x)^2 - ( \sqrt{2 - x^2} )^2 = x^2 - (2 - x^2) = x^2 - 2 + x^2 = 2x^2 - 2 = 2(x^2 - 1)$

   Therefore, the expression becomes:

   $\frac{2(x^2 - 1)}{ \left( 2x - \sqrt{2 + 2x} \right) \left( x + \sqrt{2 - x^2} \right) }$

   Notice that $x^2 - 1 = (x - 1)(x + 1)$, so we can rewrite it as:

   $\frac{2(x - 1)(x + 1)}{ \left( 2x - \sqrt{2 + 2x} \right) \left( x + \sqrt{2 - x^2} \right) }$

3. Simplify the expression and apply the limit:

   We can now proceed by further rationalizing the denominator in a similar manner if necessary.