`lim_(x to 0) (sin x + cos 3x)^(2//x)` =

Options

A. 1
B. `e^2`
C. `e^(2//3)`
D. does not exist

Explanation

To solve the limit \( \lim_{x \to 0} ( \sin x + \cos 3x )^{\frac{2}{x}} \), we will follow these steps:

### Step 1: Define the limit
Let:
\[ y = \lim_{x \to 0} ( \sin x + \cos 3x )^{\frac{2}{x}} \]

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides, we have:
\[ \ln y = \lim_{x \to 0} \frac{2}{x} \ln( \sin x + \cos 3x ) \]

### Step 3: Analyze the limit
As \( x \to 0 \):
- \( \sin x \to 0 \)
- \( \cos 3x \to 1 \)

Thus:
\[ \sin x + \cos 3x \to 0 + 1 = 1 \]
This means \( \ln( \sin x + \cos 3x ) \to \ln(1) = 0 \). Therefore, we have a \( \frac{0}{0} \) form, which allows us to apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
We need to differentiate the numerator and denominator:
\[ \ln y = \lim_{x \to 0} \frac{2 \ln( \sin x + \cos 3x )}{x} \]

Differentiating the numerator:
\[ \text{Numerator: } \frac{d}{dx} [2 \ln( \sin x + \cos 3x )] = 2 \cdot \frac{1}{\sin x + \cos 3x} \cdot (\cos x - 3 \sin 3x) \]
Differentiating the denominator:
\[ \text{Denominator: } \frac{d}{dx} [x] = 1 \]

Thus, applying L'Hôpital's Rule gives:
\[ \ln y = \lim_{x \to 0} \frac{2(\cos x - 3 \sin 3x)}{\sin x + \cos 3x} \]

### Step 5: Evaluate the limit
Now we evaluate this limit as \( x \to 0 \):
- \( \cos(0) = 1 \)
- \( \sin(0) = 0 \)
- \( \cos(3 \cdot 0) = 1 \)
- \( \sin(3 \cdot 0) = 0 \)

Substituting these values:
\[ \ln y = \frac{2(1 - 0)}{0 + 1} = 2 \]

### Step 6: Solve for \( y \)
Now we can solve for \( y \):
\[ \ln y = 2 \implies y = e^2 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} ( \sin x + \cos 3x )^{\frac{2}{x}} = e^2 \]