The expression in the image asks to evaluate the following limit:

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

Step-by-Step Solution:

1. Consider the limit as $x \to \infty$:

   • As $x \to \infty$, $\frac{10}{x} \to 0$. Therefore, the expression inside the square root becomes: $\lim_{x \to \infty} \sqrt{9 + \frac{10}{x}} = \sqrt{9 + 0} = 3$ However, the full expression involves $\sqrt{9 + \frac{10}{x}} - 3$, which seems to approach $0$ as $x$ grows larger.

2. Apply an approximation: For large values of $x$, we can expand the square root using a binomial expansion around $\frac{10}{x}$ being small: $\sqrt{9 + \frac{10}{x}} \approx 3 + \frac{1}{2} \cdot \frac{10}{x \cdot 3} = 3 + \frac{5}{3x}$

3. Substitute the approximation: Now substitute this back into the original expression: $2x \left( \sqrt{9 + \frac{10}{x}} - 3 \right) \approx 2x \left( \left( 3 + \frac{5}{3x} \right) - 3 \right) = 2x \left( \frac{5}{3x} \right)$

4. Simplify: $2x \cdot \frac{5}{3x} = \frac{10}{3}$

Thus, the value of the limit is:

$\boxed{\frac{10}{3}}$

Would you like further explanation or details on any step?

Here are 5 related questions to expand your understanding:

1. What is a binomial expansion, and how is it used for approximations?
2. How do limits behave when involving square roots as $x \to \infty$?
3. Can you apply similar methods to solve limits involving other types of radicals?
4. What happens if you change the constants inside the square root—how does it affect the limit?
5. How would this problem change if the term was $\frac{10}{x^2}$ instead of $\frac{10}{x}$?

Tip: When faced with complex square roots in limits, approximation techniques like binomial expansion help simplify the problem for large or small $x$ values.