The value of `lim_(xrarroo)(e^(x+1)log(x^(3)e^(-x)+1))/(sin^(3)(2x))` is equal to
(Use e = 2.7)

Answer

To solve the limit \( \lim_{x \to \infty} \frac{e^{x+1} \log(x^3 e^{-x} + 1)}{\sin^3(2x)} \), we will analyze the behavior of the numerator and denominator as \( x \) approaches infinity.

### Step-by-Step Solution:

1. **Analyze the Numerator**:
The numerator is \( e^{x+1} \log(x^3 e^{-x} + 1) \).
- As \( x \to \infty \), \( e^{x+1} \) grows exponentially.
- Now, consider \( \log(x^3 e^{-x} + 1) \):
\[ x^3 e^{-x} = \frac{x^3}{e^x} \]
As \( x \to \infty \), \( e^x \) grows much faster than \( x^3 \), thus \( \frac{x^3}{e^x} \to 0 \).
Therefore, \( x^3 e^{-x} + 1 \to 1 \) and \( \log(x^3 e^{-x} + 1) \to \log(1) = 0 \).

2. **Combine the Results**:
- The numerator \( e^{x+1} \log(x^3 e^{-x} + 1) \) behaves like \( e^{x+1} \cdot 0 = 0 \) as \( x \to \infty \).

3. **Analyze the Denominator**:
The denominator is \( \sin^3(2x) \).
- The sine function oscillates between -1 and 1 for all real \( x \). Thus, \( \sin^3(2x) \) will oscillate between -1 and 1, and it will not approach 0 as \( x \to \infty \).

4. **Form of the Limit**:
Now we have:
\[ \text{Numerator} \to 0 \quad \text{and} \quad \text{Denominator} \text{ oscillates between -1 and 1}. \]
Therefore, we have a limit of the form \( \frac{0}{\text{bounded}} \).

5. **Evaluate the Limit**:
Since the numerator approaches 0 while the denominator remains bounded (not approaching 0), we conclude:
\[ \lim_{x \to \infty} \frac{0}{\text{bounded}} = 0. \]

### Final Answer:
Thus, the value of the limit is:
\[ \boxed{0}. \]