Using L' Hospital's Rule evaluate :
`Lim_(x to o)(1+sinx)^(cotx)`

Answer

To evaluate the limit \( \lim_{x \to 0} (1 + \sin x)^{\cot x} \) using L'Hôpital's Rule, we can follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ L = \lim_{x \to 0} (1 + \sin x)^{\cot x} \]
This expression is of the form \( 1^\infty \) as \( x \to 0 \). To handle this, we can take the natural logarithm of \( L \):
\[ \ln L = \lim_{x \to 0} \cot x \cdot \ln(1 + \sin x) \]

### Step 2: Analyze the Form
As \( x \to 0 \), \( \cot x \) approaches \( \infty \) and \( \ln(1 + \sin x) \) approaches \( \ln(1 + 0) = 0 \). Thus, we have an indeterminate form \( \infty \cdot 0 \). We can rewrite this as:
\[ \ln L = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{\tan x} \]
Now, this is in the form \( \frac{0}{0} \), which allows us to apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
We differentiate the numerator and the denominator:
- The derivative of the numerator \( \ln(1 + \sin x) \) is:
\[ \frac{d}{dx} \ln(1 + \sin x) = \frac{\cos x}{1 + \sin x} \]
- The derivative of the denominator \( \tan x \) is:
\[ \frac{d}{dx} \tan x = \sec^2 x \]
Now we apply L'Hôpital's Rule:
\[ \ln L = \lim_{x \to 0} \frac{\frac{\cos x}{1 + \sin x}}{\sec^2 x} = \lim_{x \to 0} \frac{\cos x \cdot \cos^2 x}{1 + \sin x} \]

### Step 4: Simplify the Expression
This simplifies to:
\[ \ln L = \lim_{x \to 0} \frac{\cos^3 x}{1 + \sin x} \]
Now, substituting \( x = 0 \):
\[ \ln L = \frac{\cos^3(0)}{1 + \sin(0)} = \frac{1^3}{1 + 0} = 1 \]

### Step 5: Exponentiate to Find L
Since \( \ln L = 1 \), we exponentiate to find \( L \):
\[ L = e^1 = e \]

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