\( \operatorname { im } _ { k \rightarrow 0 } ( \frac { 1 } { 2 x } - \frac { 2 } { \sin x } \)

It appears you are interested in evaluating the limit: \[ \lim_{x \to 0} \left( \frac{1}{2x} - \frac{2}{\sin x} \right) \] Let's analyze this step by step. ### Step 1: Expand \(\sin x\) Using Taylor Series For small \(x\), the sine function can be approximated by its Taylor series expansion: \[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \] ### Step 2: Substitute the Expansion into the Expression Substituting the expansion into \(\frac{2}{\sin x}\): \[ \frac{2}{\sin x} \approx \frac{2}{x - \frac{x^3}{6} + \cdots} = \frac{2}{x\left(1 - \frac{x^2}{6} + \cdots\right)} = \frac{2}{x} \left(1 + \frac{x^2}{6} + \cdots\right) \] ### Step 3: Rewrite the Original Expression Now, substitute the approximation back into the original limit: \[ \frac{1}{2x} - \frac{2}{\sin x} \approx \frac{1}{2x} - \frac{2}{x} \left(1 + \frac{x^2}{6}\right) = \frac{1}{2x} - \frac{2}{x} - \frac{x}{3} \] Simplifying further: \[ \frac{1}{2x} - \frac{2}{x} = -\frac{3}{2x} \] Thus, the expression becomes: \[ -\frac{3}{2x} - \frac{x}{3} \] ### Step 4: Evaluate the Limit As \(x \to 0\): \[ -\frac{3}{2x} \to -\infty \quad \text{and} \quad -\frac{x}{3} \to 0 \] Therefore, the dominant term is \(-\frac{3}{2x}\), which tends to \(-\infty\). ### Conclusion \[ \lim_{x \to 0} \left( \frac{1}{2x} - \frac{2}{\sin x} \right) = -\infty \] **Answer:** The limit is minus infinity.