\( \lim _ { x \rightarrow . ^ { - } } \frac { \sqrt { 1 - \sqrt { 1 + x } } } { x } \)

To evaluate the limit \( \lim_{x \rightarrow 0^{-}} \frac{\sqrt{1 - \sqrt{1 + x}}}{x} \), we start by recognizing that as \( x \) approaches \( 0 \) from the left, \( \sqrt{1 + x} \) approaches \( 1 \). Hence, we can look into simplifying the expression. First, calculate \( \sqrt{1 + x} \). As \( x \to 0^{-} \), we can use the binomial approximation for small values: \[ \sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8} + \ldots \] Thus, \[ 1 - \sqrt{1 + x} \approx 1 - \left( 1 + \frac{x}{2} - \frac{x^2}{8} \right) = -\frac{x}{2} + \frac{x^2}{8} + O(x^3). \] Now, substituting this approximation into \( \sqrt{1 - \sqrt{1 + x}} \): \[ \sqrt{1 - \sqrt{1 + x}} \approx \sqrt{-\frac{x}{2} + \frac{x^2}{8}}. \] For small \( x \), we can take the leading term: \[ \sqrt{1 - \sqrt{1 + x}} \approx \sqrt{-\frac{x}{2}} = i\sqrt{\frac{-x}{2}} \text{ because } x \text{ is negative as } x \to 0^{-}. \] Thus, we have: \[ \sqrt{1 - \sqrt{1 + x}} \approx i \cdot \sqrt{\frac{-x}{2}} = \frac{i \sqrt{-x}}{\sqrt{2}}. \] Substituting this back into the limit, \[ \lim_{x \rightarrow 0^{-}} \frac{\frac{i \sqrt{-x}}{\sqrt{2}}}{x} = \lim_{x \rightarrow 0^{-}} \frac{i}{\sqrt{2}} \cdot \frac{\sqrt{-x}}{x} = \lim_{x \rightarrow 0^{-}} \frac{i}{\sqrt{2}} \cdot \frac{1}{\sqrt{-x}}. \] As \( x \) approaches \( 0^{-} \), \( \sqrt{-x} \) approaches \( 0 \), causing \( \frac{1}{\sqrt{-x}} \) to go to infinity. Hence the limit diverges to: \[ \lim_{x \rightarrow 0^{-}} \frac{i}{\sqrt{2}} \cdot \infty = -\infty. \] Thus, the limit is: \[ \boxed{-\infty}. \]