Evaluate `lim_(xto0) (cot2x-cosec2x)/(x).`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{\cot(2x) - \csc(2x)}{x}, \]

we will follow these steps:

### Step 1: Rewrite cotangent and cosecant
We start by rewriting \(\cot(2x)\) and \(\csc(2x)\) in terms of sine and cosine:

\[ \cot(2x) = \frac{\cos(2x)}{\sin(2x)} \quad \text{and} \quad \csc(2x) = \frac{1}{\sin(2x)}. \]

Thus, we can rewrite the expression as:

\[ \lim_{x \to 0} \frac{\frac{\cos(2x)}{\sin(2x)} - \frac{1}{\sin(2x)}}{x} = \lim_{x \to 0} \frac{\cos(2x) - 1}{x \sin(2x)}. \]

### Step 2: Factor out the limit
Now, we can separate the limit:

\[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x} \cdot \frac{1}{\sin(2x)}. \]

### Step 3: Use known limits
We know that

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{(2x)^2} = \frac{1}{2}, \]

which implies

\[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x^2} = -\frac{1}{2}. \]

Thus,

\[ \lim_{x \to 0} \frac{\cos(2x) - 1}{x} = \lim_{x \to 0} \frac{\cos(2x) - 1}{(2x)^2} \cdot 2x = -\frac{1}{2} \cdot 2x = -x. \]

### Step 4: Substitute into the limit
Now substituting back, we have:

\[ \lim_{x \to 0} \frac{-x}{x \sin(2x)} = \lim_{x \to 0} \frac{-1}{\sin(2x)}. \]

### Step 5: Evaluate the limit
As \(x \to 0\), \(\sin(2x) \to 2x\), so we have:

\[ \lim_{x \to 0} \frac{-1}{\sin(2x)} = \lim_{x \to 0} \frac{-1}{2x} = -\frac{1}{2}. \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{\cot(2x) - \csc(2x)}{x} = -1. \]