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The given problem asks to find the limit:

\[
\lim_{{x \to \infty}} \frac{x^3}{\sqrt{16x^9 + 11}}
\]

### Step-by-step solution:

1. **Simplifying the expression**:
   \[
   \lim_{{x \to \infty}} \frac{x^3}{\sqrt{16x^9 + 11}}
   \]

   Notice that as \(x \to \infty\), the term \(16x^9\) dominates over the constant 11. Therefore, for large values of \(x\), the expression simplifies to:

   \[
   \frac{x^3}{\sqrt{16x^9 + 11}} \approx \frac{x^3}{\sqrt{16x^9}}
   \]

2. **Simplifying the square root**:
   \[
   \sqrt{16x^9} = 4x^{9/2}
   \]
   Therefore, the limit becomes:
   \[
   \lim_{{x \to \infty}} \frac{x^3}{4x^{9/2}} = \lim_{{x \to \infty}} \frac{1}{4x^{3/2}}
   \]

3. **Evaluating the limit**:
   As \(x \to \infty\), the term \(4x^{3/2}\) grows without bound, so:
   \[
   \lim_{{x \to \infty}} \frac{1}{4x^{3/2}} = 0
   \]

Thus, the limit is:

\[
\boxed{0}
\]

### Do you have any questions or would you like further explanation?

### Related questions:
1. What are some key strategies to solve limits at infinity?
2. How does the behavior of polynomials and roots affect limits as \(x\) approaches infinity?
3. How do we handle limits where the dominant term grows faster than others?
4. Can limits at infinity always be simplified using dominant terms?
5. What happens when the powers in the numerator and denominator are equal at infinity?

**Tip:** When dealing with limits at infinity, always focus on the highest powers in the numerator and denominator to simplify the calculation.

Find the following limit. If the limit does not exist, enter DNE. 

\(\lim_{x\to\infty} \frac{x^3}{\sqrt{16x^9 + 11}}\)

Explore the process of finding the limit of the function \(\frac{x^3}{\sqrt{16x^9 + 11}}\) as x approaches infinity. This solution employs asymptotic behavior and radical simplification to evaluate the limit, suitable for college or university level students.

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