We are given the limit expression:

$\lim_{x \to \infty} x \left( \sqrt{4 + \frac{1}{x}} - 2 \right)$

Step-by-Step Solution:

1. Simplify the expression inside the square root: As $x \to \infty$, the term $\frac{1}{x}$ approaches 0. So, we can analyze the behavior of the square root term: $\sqrt{4 + \frac{1}{x}} \to \sqrt{4} = 2$ However, we need to calculate the behavior of the expression more carefully for large values of $x$, so we'll expand it using a binomial approximation.

2. Use binomial approximation: For large $x$, we can use the binomial approximation for square roots: $\sqrt{1 + z} \approx 1 + \frac{z}{2} \quad \text{for small } z$ Here, $z = \frac{1}{4x}$, so: $\sqrt{4 + \frac{1}{x}} = 2 \sqrt{1 + \frac{1}{4x}} \approx 2 \left(1 + \frac{1}{8x}\right) = 2 + \frac{1}{4x}$

3. Substitute into the original expression: Now substitute this approximation into the limit expression: $x \left( \sqrt{4 + \frac{1}{x}} - 2 \right) \approx x \left( 2 + \frac{1}{4x} - 2 \right) = x \left( \frac{1}{4x} \right)$ Simplifying further: $x \times \frac{1}{4x} = \frac{1}{4}$

4. Take the limit: Since the expression simplifies to a constant value $\frac{1}{4}$, the limit is: $\lim_{x \to \infty} x \left( \sqrt{4 + \frac{1}{x}} - 2 \right) = \frac{1}{4}$

Final Answer:

$\boxed{\frac{1}{4}}$

Would you like further details on any step, or have more questions? Here are some related problems you could explore:

1. What if the limit was $\lim_{x \to 0} x \left( \sqrt{4 + \frac{1}{x}} - 2 \right)$?
2. How does the binomial approximation apply in different contexts for small values of $z$?
3. What would happen if the term inside the square root was more complex, like $\sqrt{4 + \frac{1}{x^2}}$?
4. Can you find an alternative method for solving this limit, such as L'Hopital's Rule?
5. How can we use Taylor series to approximate square roots more generally?

Tip: Binomial expansions are very useful for approximating functions for large or small values of variables. Familiarize yourself with these expansions for common functions!