We are tasked with finding the limit:

$\lim_{x \to \infty} \frac{1}{2x - \sqrt{4x^2 - 5x + 6}}.$

Step-by-Step Solution:

We begin by simplifying the expression inside the limit.

1. Simplify the denominator: The expression inside the denominator is $2x - \sqrt{4x^2 - 5x + 6}$. To understand the behavior as $x \to \infty$, we focus on the dominant terms as $x$ grows large.

   The quadratic expression $4x^2 - 5x + 6$ can be approximated for large $x$ as: $4x^2 - 5x + 6 \approx 4x^2.$ Therefore, the square root of this expression behaves like: $\sqrt{4x^2 - 5x + 6} \approx \sqrt{4x^2} = 2x.$ So, the denominator approximates to: $2x - \sqrt{4x^2 - 5x + 6} \approx 2x - 2x = 0.$

   However, this suggests that the denominator is approaching 0, but we need more precision to understand how it behaves as $x \to \infty$.

2. Refine the denominator: To refine the approximation, we factor out $x^2$ from the square root to see the next-order behavior: $\sqrt{4x^2 - 5x + 6} = \sqrt{4x^2\left(1 - \frac{5}{4x} + \frac{6}{4x^2}\right)} = 2x \sqrt{1 - \frac{5}{4x} + \frac{6}{4x^2}}.$ For large $x$, we can use the binomial approximation $\sqrt{1 + z} \approx 1 + \frac{z}{2}$ for small $z$, where $z = -\frac{5}{4x} + \frac{6}{4x^2}$. Thus, we have: $\sqrt{1 - \frac{5}{4x} + \frac{6}{4x^2}} \approx 1 - \frac{5}{8x}.$ Therefore, the denominator becomes: $2x - 2x\left(1 - \frac{5}{8x}\right) = 2x - 2x + \frac{5}{4} = \frac{5}{4}.$

3. Evaluate the limit: Now, we can evaluate the limit: $\lim_{x \to \infty} \frac{1}{2x - \sqrt{4x^2 - 5x + 6}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}.$

Final Answer:

$\lim_{x \to \infty} \frac{1}{2x - \sqrt{4x^2 - 5x + 6}} = \frac{4}{5}.$

Would you like more details or further clarifications? Here are some related questions for further exploration:

1. What is the behavior of the expression as $x$ approaches negative infinity?
2. How does the binomial expansion apply to approximating expressions like this?
3. Can this method of simplifying the denominator be applied to other limits involving square roots?
4. What happens if we replace the quadratic expression with a cubic one? How does it affect the limit?
5. How would the limit change if there were a higher-order term in the denominator?

Tip: When handling limits involving square roots and large values of $x$, always try to identify the dominant terms and simplify them first for a more manageable approximation.