The value of `lim_(xrarr0)((4^x-1)^3)/(sin.(x^2)/(4)log(1+3x))`,is

Options

A. `(4)/(3)(In 4)^2`
B. `(4)/(3)(In 4)^3`
C. `(3)/(2)(In 4)^2`
D. `(3)/(2)(In 4)^3`

Explanation

To solve the limit \( \lim_{x \to 0} \frac{(4^x - 1)^3}{\frac{\sin(x^2)}{4} \log(1 + 3x)} \), we will use some fundamental limit properties. Let's break it down step by step.

### Step 1: Rewrite the limit
We start with the expression:
\[ \lim_{x \to 0} \frac{(4^x - 1)^3}{\frac{\sin(x^2)}{4} \log(1 + 3x)} \]
We can rewrite this as:
\[ \lim_{x \to 0} \frac{(4^x - 1)^3}{\sin(x^2) \cdot \frac{1}{4} \log(1 + 3x)} = 4 \cdot \lim_{x \to 0} \frac{(4^x - 1)^3}{\sin(x^2) \log(1 + 3x)} \]

### Step 2: Use limit properties
We know the following limits:
1. \( \lim_{x \to 0} \frac{4^x - 1}{x} = \ln(4) \)
2. \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \)
3. \( \lim_{x \to 0} \frac{\log(1 + x)}{x} = 1 \)

### Step 3: Rewrite \( (4^x - 1)^3 \)
Using the first property, we can express \( (4^x - 1) \) in terms of \( x \):
\[ 4^x - 1 = x \cdot \frac{4^x - 1}{x} \to x \cdot \ln(4) \text{ as } x \to 0 \]
Thus,
\[ (4^x - 1)^3 = (x \cdot \ln(4))^3 = x^3 \cdot (\ln(4))^3 \]

### Step 4: Rewrite \( \sin(x^2) \)
Using the second property, we have:
\[ \sin(x^2) \approx x^2 \text{ as } x \to 0 \]

### Step 5: Rewrite \( \log(1 + 3x) \)
Using the third property, we find:
\[ \log(1 + 3x) \approx 3x \text{ as } x \to 0 \]

### Step 6: Substitute back into the limit
Now substituting these approximations back into the limit:
\[ \lim_{x \to 0} \frac{x^3 (\ln(4))^3}{x^2 \cdot 3x} = \lim_{x \to 0} \frac{x^3 (\ln(4))^3}{3x^3} = \frac{(\ln(4))^3}{3} \]

### Step 7: Multiply by 4
Finally, remember we factored out a 4 earlier:
\[ 4 \cdot \frac{(\ln(4))^3}{3} = \frac{4(\ln(4))^3}{3} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{4(\ln(4))^3}{3} \]