We are tasked to evaluate the limit as $x \to \infty$ for the given expression:

$\lim_{x \to \infty} \frac{\sqrt{30x^3 - 15x^2 + 16x^6}}{7x^2 + 7}.$

Step 1: Simplify the square root expression

In the numerator, factor out $x^6$ from the terms inside the square root: $\sqrt{30x^3 - 15x^2 + 16x^6} = \sqrt{x^6 \left( \frac{30}{x^3} - \frac{15}{x^4} + 16 \right)}.$ This simplifies to: $x^3 \sqrt{\frac{30}{x^3} - \frac{15}{x^4} + 16}.$

Step 2: Simplify the denominator

The denominator is: $7x^2 + 7.$

Step 3: Express the limit

The expression for the limit becomes: $\lim_{x \to \infty} \frac{x^3 \sqrt{\frac{30}{x^3} - \frac{15}{x^4} + 16}}{7x^2 + 7}.$

Step 4: Analyze the dominant terms as $x \to \infty$

1. As $x \to \infty$, the terms $\frac{30}{x^3}$ and $\frac{15}{x^4}$ inside the square root vanish. Thus, the square root simplifies to: $\sqrt{\frac{30}{x^3} - \frac{15}{x^4} + 16} \to \sqrt{16} = 4.$ So, the numerator behaves like $x^3 \cdot 4 = 4x^3$.

2. In the denominator, the dominant term is $7x^2$ since $7$ is negligible as $x \to \infty$.

Thus, the limit becomes: $\lim_{x \to \infty} \frac{4x^3}{7x^2} = \lim_{x \to \infty} \frac{4}{7}x = \infty.$

Step 5: Conclusion

The limit is infinite, so we conclude: $\text{The limit does not exist (DNE).}$

Would you like further explanation? Here are related questions you might explore:

1. How do square roots affect dominant term analysis in limits?
2. What happens if the numerator and denominator grow at the same rate?
3. How do limits involving polynomials compare with exponential functions?
4. How do you evaluate limits at $x \to -\infty$?
5. Why do we ignore smaller-order terms when $x \to \infty$?

Tip: Always factor out the highest power of $x$ inside square roots to simplify the analysis of dominant terms.