`Lt_(x to0)(log(1+3x))/(sin 4x)` is equal to

Options

A. `(4)/(3)`
B. `-(4)/(3)`
C. `(3)/(4)`
D. `-(3)/(4)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\log(1 + 3x)}{\sin(4x)} \), we will follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ L = \lim_{x \to 0} \frac{\log(1 + 3x)}{\sin(4x)} \]

### Step 2: Use the Properties of Logarithms
We can use the property of logarithms that states:
\[ \lim_{x \to 0} \frac{\log(1 + u)}{u} = 1 \quad \text{as } u \to 0 \]
In our case, let \( u = 3x \). Then as \( x \to 0 \), \( u \to 0 \).

### Step 3: Rewrite the Limit
We can rewrite the limit as:
\[ L = \lim_{x \to 0} \frac{\log(1 + 3x)}{3x} \cdot \frac{3x}{\sin(4x)} \]

### Step 4: Separate the Limits
Now we can separate the limit into two parts:
\[ L = \lim_{x \to 0} \frac{\log(1 + 3x)}{3x} \cdot \lim_{x \to 0} \frac{3x}{\sin(4x)} \]

### Step 5: Evaluate the First Limit
Using the property of logarithms:
\[ \lim_{x \to 0} \frac{\log(1 + 3x)}{3x} = 1 \]

### Step 6: Evaluate the Second Limit
For the second limit, we can use the property:
\[ \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \quad \text{for any constant } k \]
Thus,
\[ \lim_{x \to 0} \frac{3x}{\sin(4x)} = \lim_{x \to 0} \frac{3}{4} \cdot \frac{4x}{\sin(4x)} = \frac{3}{4} \cdot 1 = \frac{3}{4} \]

### Step 7: Combine the Results
Now, we can combine the results from the two limits:
\[ L = 1 \cdot \frac{3}{4} = \frac{3}{4} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\log(1 + 3x)}{\sin(4x)} = \frac{3}{4} \]