The problem you're working on involves calculating the limit:

$\lim_{x \to \infty} 2x \left( \sqrt{9 + \frac{10}{x}} - 3 \right)$

Step-by-step solution:

1. Consider the limit behavior as $x \to \infty$. The expression inside the square root can be simplified:

   $\sqrt{9 + \frac{10}{x}}$ As $x \to \infty$, $\frac{10}{x} \to 0$, so:

   $\sqrt{9 + \frac{10}{x}} \to \sqrt{9} = 3$ Thus, the expression becomes close to $3 - 3 = 0$, but to handle the limit more carefully, we need to expand the square root.

2. Apply a binomial expansion for the square root for large $x$:

   $\sqrt{9 + \frac{10}{x}} = 3 + \frac{1}{2 \cdot 3} \cdot \frac{10}{x} + O\left(\frac{1}{x^2}\right)$ This simplifies to:

   $\sqrt{9 + \frac{10}{x}} \approx 3 + \frac{5}{3x}$

3. Substitute back into the original expression:

   $2x \left( \left(3 + \frac{5}{3x}\right) - 3 \right) = 2x \cdot \frac{5}{3x}$ Simplifying this gives:

   $2x \cdot \frac{5}{3x} = \frac{10}{3}$

4. Conclusion:

   The value of the limit is $\frac{10}{3}$, and expressed with three decimal places, the result is:

   $3.333$

Final Answer:

$\boxed{3.333}$

Let me know if you would like further details or clarification!

Here are some related questions for practice:

1. What is the limit of $\lim_{x \to \infty} x \left( \sqrt{1 + \frac{1}{x}} - 1 \right)$?
2. How does the binomial expansion help in simplifying limits involving square roots?
3. What happens when you calculate $\lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)$?
4. Can you solve $\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$?
5. How can L'Hopital's Rule be applied to similar limit problems?

Tip: When dealing with limits involving square roots and infinity, the binomial expansion can be a very effective tool.