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

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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.
The limit is minus infinity.

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