`lim _(xto0)(((1+ x )^(2/x))/(e ^(2)))^((4)/(sin x))` is :

Options

A. `e ^(4)`
B. `e ^(-4)`
C. `e ^(8)`
D. `e ^(-8)`

Answer

To solve the limit problem \( \lim_{x \to 0} \left( \frac{(1+x)^{\frac{2}{x}}}{e^2} \right)^{\frac{4}{\sin x}} \), we can follow these steps:

### Step 1: Rewrite the Limit
Let \( L = \lim_{x \to 0} \left( \frac{(1+x)^{\frac{2}{x}}}{e^2} \right)^{\frac{4}{\sin x}} \).

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides gives:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \ln \left( \frac{(1+x)^{\frac{2}{x}}}{e^2} \right) \]
Using the property of logarithms, we can separate the terms:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \left( \frac{2}{x} \ln(1+x) - \ln(e^2) \right) \]
Since \( \ln(e^2) = 2 \), we have:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \left( \frac{2}{x} \ln(1+x) - 2 \right) \]

### Step 3: Simplify the Expression
This can be rewritten as:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \left( \frac{2 \ln(1+x) - 2x}{x} \right) \]

### Step 4: Apply L'Hôpital's Rule
As \( x \to 0 \), both the numerator and denominator approach 0, allowing us to apply L'Hôpital's Rule:
\[ \ln L = 8 \lim_{x \to 0} \frac{\ln(1+x) - x}{x^2} \]

### Step 5: Differentiate Numerator and Denominator
Differentiate the numerator and denominator:
- The derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \).
- The derivative of \( x \) is \( 1 \).

Applying L'Hôpital's Rule:
\[ \ln L = 8 \lim_{x \to 0} \frac{\frac{1}{1+x} - 1}{2x} \]
This simplifies to:
\[ \ln L = 8 \lim_{x \to 0} \frac{-x}{2x(1+x)} = -4 \lim_{x \to 0} \frac{1}{1+x} = -4 \]

### Step 6: Solve for L
Now we have:
\[ \ln L = -4 \implies L = e^{-4} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \left( \frac{(1+x)^{\frac{2}{x}}}{e^2} \right)^{\frac{4}{\sin x}} = e^{-4} \]