`lim_(xto0)(((1+x)^(1//x))/e)^(1/(sinx))` is equal to

Options

A. `sqrt(2)`
B. `e`
C. `1/(sqrt(e))`
D. `1/e`

Answer

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

### Step 1: Simplify the expression inside the limit
We know that as \( x \to 0 \), \( (1+x)^{\frac{1}{x}} \) approaches \( e \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{\sin x}} = \lim_{x \to 0} \left( \frac{e}{e} \right)^{\frac{1}{\sin x}} = \lim_{x \to 0} 1^{\frac{1}{\sin x}} = 1 \]

### Step 2: Identify the indeterminate form
However, we need to analyze the expression more closely since \( \frac{(1+x)^{\frac{1}{x}}}{e} \) approaches 1, leading to the indeterminate form \( 1^{\infty} \) as \( x \to 0 \).

### Step 3: Use the exponential limit form
To resolve the indeterminate form \( 1^{\infty} \), we can use the exponential limit:
\[ \lim_{x \to 0} f(x)^{g(x)} = e^{\lim_{x \to 0} g(x) \cdot (f(x) - 1)} \]
where \( f(x) = \frac{(1+x)^{\frac{1}{x}}}{e} \) and \( g(x) = \frac{1}{\sin x} \).

### Step 4: Calculate \( f(x) - 1 \)
We have:
\[ f(x) = \frac{(1+x)^{\frac{1}{x}}}{e} \]
Using the expansion \( (1+x)^{\frac{1}{x}} \approx e \left( 1 - \frac{x}{2} + O(x^2) \right) \) as \( x \to 0 \), we get:
\[ f(x) - 1 \approx \frac{e \left( 1 - \frac{x}{2} \right)}{e} - 1 = -\frac{x}{2} + O(x^2) \]

### Step 5: Substitute into the limit
Now, substituting into the limit:
\[ \lim_{x \to 0} g(x) \cdot (f(x) - 1) = \lim_{x \to 0} \frac{1}{\sin x} \cdot \left(-\frac{x}{2} + O(x^2)\right) \]
As \( x \to 0 \), \( \sin x \approx x \), thus:
\[ \lim_{x \to 0} \frac{-\frac{x}{2}}{x} = \lim_{x \to 0} -\frac{1}{2} = -\frac{1}{2} \]

### Step 6: Final result
Now we can substitute back into the exponential form:
\[ \lim_{x \to 0} f(x)^{g(x)} = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

Thus, the final answer is:
\[ \lim_{x \to 0} \left( \frac{(1+x)^{\frac{1}{x}}}{e} \right)^{\frac{1}{\sin x}} = \frac{1}{\sqrt{e}} \]