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

### Step 1: Rewrite the limit
We start by rewriting the limit in a more manageable form:
\[ L = \lim_{x \to 0} \left( \frac{(1 + x)^{\frac{2}{x}}}{e^2} \right)^{\frac{4}{\sin x}} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \ln \left( \frac{(1 + x)^{\frac{2}{x}}}{e^2} \right) \]
Using the properties of logarithms, this can be simplified to:
\[ \ln L = \lim_{x \to 0} \frac{4}{\sin x} \left( \frac{2}{x} \ln(1 + x) - 2 \right) \]

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

### Step 4: Evaluate the limit
As \( x \to 0 \), we know that \( \ln(1 + x) \approx x - \frac{x^2}{2} + O(x^3) \). Therefore,
\[ \frac{\ln(1 + x)}{x} \to 1 \text{ as } x \to 0 \]
Thus,
\[ \frac{2 \ln(1 + x)}{x} \to 2 \text{ as } x \to 0 \]
This means:
\[ \frac{2 \ln(1 + x)}{x} - 2 \to 0 \text{ as } x \to 0 \]
Since \( \sin x \approx x \) as \( x \to 0 \), we can replace \( \sin x \) with \( x \):
\[ \ln L = \lim_{x \to 0} \frac{4}{x} \cdot 0 = 0 \]

### Step 5: Exponentiate to find L
Since \( \ln L = 0 \), we have:
\[ L = e^0 = 1 \]

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