Evaluate: `lim_(x to 0)(cotx)^(sin2x)`

Answer

To evaluate the limit \( \lim_{x \to 0} (\cot x)^{\sin 2x} \), we can follow these steps:

### Step 1: Define the Limit
Let \( L = \lim_{x \to 0} (\cot x)^{\sin 2x} \).

### Step 2: Take the Natural Logarithm
Taking the natural logarithm on both sides gives:
\[ \log L = \lim_{x \to 0} \sin 2x \cdot \log(\cot x) \]

### Step 3: Analyze the Limit
As \( x \to 0 \), we know:
- \( \cot x \to \infty \) (since \( \cot x = \frac{\cos x}{\sin x} \))
- \( \sin 2x \to 0 \)

This results in the form \( 0 \cdot \infty \). To resolve this, we can rewrite \( \sin 2x \) in terms of \( \cot x \).

### Step 4: Rewrite \( \sin 2x \)
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can express the limit as:
\[ \log L = \lim_{x \to 0} 2 \sin x \cos x \cdot \log(\cot x) \]

### Step 5: Change the Form
We can use the fact that \( \cot x = \frac{\cos x}{\sin x} \), so:
\[ \log(\cot x) = \log(\cos x) - \log(\sin x) \]
Thus, we have:
\[ \log L = \lim_{x \to 0} 2 \sin x \cos x \cdot (\log(\cos x) - \log(\sin x)) \]

### Step 6: Evaluate Each Logarithm
As \( x \to 0 \):
- \( \log(\cos x) \to \log(1) = 0 \)
- \( \log(\sin x) \to \log(x) \to -\infty \)

This gives us the form \( 0 \cdot (-\infty) \) again. We can focus on \( -\log(\sin x) \).

### Step 7: Apply L'Hôpital's Rule
We can rewrite the limit as:
\[ \log L = \lim_{x \to 0} \frac{2 \sin x \log(\cot x)}{\frac{1}{\sin 2x}} \]
Now, we can apply L'Hôpital's rule since it is of the form \( \frac{0}{0} \).

### Step 8: Differentiate Numerator and Denominator
Differentiate the numerator and denominator:
- The derivative of the numerator \( 2 \sin x \log(\cot x) \) involves product and chain rules.
- The derivative of the denominator \( \frac{1}{\sin 2x} \) is \( -2 \cos 2x \).

### Step 9: Simplify and Evaluate the Limit
After applying L'Hôpital's rule and simplifying, we find:
\[ \log L = 0 \]
Thus, \( L = e^0 = 1 \).

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} (\cot x)^{\sin 2x} = 1 \]