\( = \lim _ { x \rightarrow 0 } \frac { 2 x - \sin x } { \sqrt { 1 - \cos x } } \)

To evaluate the limit \( \lim _ { x \rightarrow 0 } \frac { 2 x - \sin x } { \sqrt { 1 - \cos x } } \), let's start with simplifying the expression. We know that as \( x \) approaches 0, both the numerator and denominator approach 0, resulting in an indeterminate form. Using the Taylor series expansions near \( x = 0 \): 1. The sine function can be approximated as: \[ \sin x \approx x - \frac{x^3}{6} + O(x^5). \] Thus, \( 2x - \sin x \approx 2x - \left(x - \frac{x^3}{6}\right) = x + \frac{x^3}{6} \). 2. For the cosine function: \[ 1 - \cos x \approx \frac{x^2}{2} - \frac{x^4}{24} + O(x^6). \] Therefore, \( \sqrt{1 - \cos x} \approx \sqrt{\frac{x^2}{2}} = \frac{x}{\sqrt{2}} \). Now substituting these approximations into the limit: \[ \frac{2x - \sin x}{\sqrt{1 - \cos x}} \approx \frac{x + \frac{x^3}{6}}{\frac{x}{\sqrt{2}}} = \frac{\sqrt{2} \left(x + \frac{x^3}{6}\right)}{x} = \sqrt{2} \left(1 + \frac{x^2}{6}\right). \] As \( x \rightarrow 0 \), the term \( \frac{x^2}{6} \) goes to 0, leading us to: \[ \lim_{x \to 0} \sqrt{2} \left(1 + \frac{x^2}{6}\right) = \sqrt{2}. \] Hence, the limit is: \[ \lim _ { x \rightarrow 0 } \frac { 2 x - \sin x } { \sqrt { 1 - \cos x } } = \sqrt{2}. \]