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 1: Analyze the Numerator

The numerator is

\[ e^{x+1} \log(x^3 e^{-x} + 1). \]

As \( x \to \infty \), we can simplify \( x^3 e^{-x} \):

\[ x^3 e^{-x} = \frac{x^3}{e^x}. \]

Since \( e^x \) grows much faster than \( x^3 \), we have:

\[ x^3 e^{-x} \to 0 \quad \text{as } x \to \infty. \]

Thus,

\[ x^3 e^{-x} + 1 \to 1. \]

Now, we can evaluate the logarithm:

\[ \log(x^3 e^{-x} + 1) \to \log(1) = 0. \]

However, we need to find the rate at which it approaches zero. Using the approximation \( \log(1 + u) \approx u \) for small \( u \):

\[ \log(x^3 e^{-x} + 1) \approx x^3 e^{-x} \quad \text{as } x \to \infty. \]

### Step 2: Substitute Back into the Numerator

Now substituting back, we have:

\[ e^{x+1} \log(x^3 e^{-x} + 1) \approx e^{x+1} \cdot x^3 e^{-x} = e \cdot x^3. \]

### Step 3: Analyze the Denominator

The denominator is

\[ \sin^3(2x). \]

The sine function oscillates between -1 and 1, hence \( \sin^3(2x) \) will oscillate between -1 and 1 as well. Therefore, the denominator remains bounded and does not approach zero.

### Step 4: Combine the Results

Now, we can rewrite the limit:

\[ \lim_{x \to \infty} \frac{e \cdot x^3}{\sin^3(2x)}. \]

Since \( \sin^3(2x) \) is bounded, the limit will depend on the behavior of \( x^3 \):

\[ \text{As } x \to \infty, \quad \frac{e \cdot x^3}{\sin^3(2x)} \to \infty. \]

### Conclusion

Thus, the value of the limit is:

\[ \lim_{x \to \infty} \frac{e^{x+1} \log(x^3 e^{-x} + 1)}{\sin^3(2x)} = \infty. \]

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