The problem involves finding the limit:

$\lim_{x \to \infty} \left( \sqrt{5x^2 + 3x + 3} - \sqrt{6x^2 + 3x - 2} \right)$

Steps for solving:

1. Factor out the highest power of $x^2$ inside the square roots:

   • For the first term: $\sqrt{5x^2 + 3x + 3} = \sqrt{x^2(5 + \frac{3}{x} + \frac{3}{x^2})} = |x| \sqrt{5 + \frac{3}{x} + \frac{3}{x^2}}$
   • For the second term: $\sqrt{6x^2 + 3x - 2} = \sqrt{x^2(6 + \frac{3}{x} - \frac{2}{x^2})} = |x| \sqrt{6 + \frac{3}{x} - \frac{2}{x^2}}$

   As $x \to \infty$, we can treat $|x| = x$, since we are considering positive infinity.

2. Simplify for large $x$: As $x \to \infty$, the terms $\frac{3}{x}$, $\frac{3}{x^2}$, and $\frac{-2}{x^2}$ approach 0. So the expressions inside the square roots simplify to:

   $\sqrt{5 + \frac{3}{x} + \frac{3}{x^2}} \to \sqrt{5}, \quad \sqrt{6 + \frac{3}{x} - \frac{2}{x^2}} \to \sqrt{6}$

3. Take the limit: The expression now becomes:

   $\lim_{x \to \infty} \left( x\sqrt{5} - x\sqrt{6} \right)$

   Factor out $x$:

   $x\left( \sqrt{5} - \sqrt{6} \right)$

   Since $\sqrt{5} - \sqrt{6}$ is negative, the whole expression tends to $-\infty$ as $x \to \infty$.

Conclusion:

The limit is $-\infty$, which corresponds to option c.

Would you like a detailed explanation of any specific step, or further assistance?

Here are some related questions you might find useful:

1. What happens when the coefficients inside the square roots are equal?
2. How does factoring the highest degree of $x$ help in limit problems?
3. What are some techniques for handling limits involving square roots at infinity?
4. How does the behavior of functions change as $x \to -\infty$?
5. What are common mistakes to avoid when solving limits at infinity?

Tip: When solving limits at infinity involving square roots, factoring out the highest power of $x^2$ often simplifies the process.