The value of `lim_(xto0)(1)/(x) int_(0)^(x)(1+ sin 2t)^(1/t) dt` equals :

Options

A. 2
B. 1
C. e
D. `e^(2)`

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{1}{x} \int_{0}^{x} (1 + \sin 2t)^{\frac{1}{t}} dt, \]

we will follow these steps:

### Step 1: Identify the limit form
As \( x \to 0 \), the integral from \( 0 \) to \( x \) approaches \( 0 \), and thus the entire expression becomes \( \frac{0}{0} \). This indicates we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
We need to differentiate the numerator and the denominator. The denominator is simply \( x \), which differentiates to \( 1 \). For the numerator, we apply the Leibniz rule for differentiation under the integral sign:

\[ \frac{d}{dx} \int_{0}^{x} (1 + \sin 2t)^{\frac{1}{t}} dt = (1 + \sin 2x)^{\frac{1}{x}} \cdot \frac{d}{dx}(x) - 0 = (1 + \sin 2x)^{\frac{1}{x}}. \]

Thus, we have:

\[ \lim_{x \to 0} \frac{(1 + \sin 2x)^{\frac{1}{x}}}{1}. \]

### Step 3: Evaluate the limit
Now we need to evaluate

\[ \lim_{x \to 0} (1 + \sin 2x)^{\frac{1}{x}}. \]

As \( x \to 0 \), \( \sin 2x \to 0 \), leading to the form \( 1^{\infty} \). To resolve this, we take the natural logarithm:

\[ \log y = \lim_{x \to 0} \frac{1}{x} \log(1 + \sin 2x). \]

### Step 4: Apply L'Hôpital's Rule again
This expression is again in the \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again:

\[ \log y = \lim_{x \to 0} \frac{\log(1 + \sin 2x)}{x}. \]

Differentiating the numerator and denominator gives:

\[ \frac{d}{dx} \log(1 + \sin 2x) = \frac{2 \cos 2x}{1 + \sin 2x}, \]

and the denominator differentiates to \( 1 \). Thus, we have:

\[ \log y = \lim_{x \to 0} \frac{2 \cos 2x}{1 + \sin 2x}. \]

### Step 5: Evaluate the limit
Now substituting \( x = 0 \):

\[ \log y = \frac{2 \cdot \cos 0}{1 + \sin 0} = \frac{2 \cdot 1}{1 + 0} = 2. \]

### Step 6: Solve for \( y \)
Exponentiating both sides gives:

\[ y = e^2. \]

### Final Answer
Thus, the value of the limit is

\[ \lim_{x \to 0} \frac{1}{x} \int_{0}^{x} (1 + \sin 2t)^{\frac{1}{t}} dt = e^2. \]